Data Structures and Algorithms Stack in C and C++

1. Overview DataStructures&Algorithms Series1: Hi, welcome to the first session of the course, stack. Use and linked list practical programs, DSA, that is data structures and algorithms series one, using both C language as well as C plus plus. Here, the very first session of the course, I will be listing what all topics we will be covering in the entire course. You'll be understanding the data structures that are stacked use and linked list concert twice. And we will be doing lots of hands-on in order to understand these data structures. We will be also doing the hands-on in order to understand the applications of this data structure. I have covered all this lecture sessions on whiteboard as well as on both systems so that you will understand these data structures more thoroughly. So let's start and let's see what topics we will be covering in depth in this complete. Moving on to the new slide, that is Session one, different sessions. And under decisions there will be lectures. The very first session which is in production. Because structures here we will be seeing what are the cost structures, what are data structures are there, and what is the importance? What are its advantages and disadvantages to details about data structures we'll be seeing in this session, one session to this stack theoretical and practical sessions on my board. That is, we will understand that the test structured that is stacked. You'll understand this concept will see the program and they too have explained on the whiteboard solidarity video lectures under this section, lecture one and lecture two, I will be covering stack introduction. So there are small video clips. I have a word in part one and part two of the same stack and production. Didn't. We will be seeing the detail about the stack. What is stack, how it is used? Much mode in these lectures we will understand. Then in lecture three, we will be seeing the program that is stack using arrays with programming in C, the program how to implement the stack using it. Eddie. So there are different ways in which we can implement the stack. So you can see the program in this lecture three will understand in more detail. Then lecture four is to implement the stack using linked list in C, the program for the same and see how to implement stacks using linked list. Then the new session, that is session that we'll be covering the application that is stack application on by both. So I have a word here, the application of the stack I've explained on the board. So there are different lectures under this session. So an ecto 123 is one theme application that is in order to reverse the string, but using the stack. Then same session, the different application of the stack. This lecture, we will see the expression is having them at balanced parenthesis or not. So here I'll explain you. What is this balanced parenthesis? So detail about it. I will explain you in the dedicated lectures, just understand this is one important application on the stack. We achieve more efficiently using the stack. So different video clips or their part one, part two to achieve the same application. And lecture six is another application, important application of the stack that is infix to postfix conversion left to right associativity. Here in this lecture we will understand what is in fixed, what is post fixed by this conversion required and how to achieve this conversion. So partial fallen short to see this main purpose of this application is to evaluate your mathematical expression very quickly. This conversion is required and there are different ways to achieve it. Left to right associativity in fixed or fourth fixed gun motion doing left or right associativity and just have an infix to postfix conversion using doing right to left, left to right. And this is right-to-left, different minutes. So just knock one. Now just understand this is in order to evaluate your mathematical expression really quickly. This application is very much required in detail, we'll see in the dedicated lectures. Then another lecture for the same session for the application on the stack is in fixed to force the program for concept. That is how to achieve it, how to achieve and motion of infix to postfix. That results in this lecture, we will see the concept Lecture 67. And the program is, I have dedicated lecture 89. It is the same program. In order to achieve that is, the program will be the same. Asked that isn't fixed to cause fixed fun version. You'll see the program in this lecture 89 and nature and spot evaluation of the course, fixed postfix evaluation. So that means here you can see that finally, we will get the postfix conversion. So we have to evaluate that expression, which will be in the postfix form that we will see the concept in lecture ten and we will see the program in Lecture 11. In, in order to evaluate the course with all the details about this, maybe you've seen the dedicated lectures. These are the important application on the stack. Then. Dialect known. Then the another lecture dwell that is for the same session for the stack application is in fixed or prefix conversion rule with examples. So earlier sessions lectures were called infix to postfix. This lecture is in fixed to prefix. This same posts on this fun version again to say it is to evaluate the mathematical expression quickly. This you can also convert infix to prefix. So you can do in fixture costs fixed, you can do in Pix2Pix also be the same. Now in order to evaluate the information quickly are required for the details about it. You'll see in electric drill and we will do the recording, will write the program, feed the same in fixture group prefects in lecture talking. This is all about the application of stack, the new session in this stack theoretical and practical session on left pump that is on the system. So we will be writing the program and executing on the operating system, on the system using C and C plus plus, we will see different programs of the stack and we'll execute lecture one is to write the program and execute to implement stack using an array in C language. And C plus plus will implement the stack using erin write a program for it and will execute it if you see language and C plus plus. Lecture two is to implement the stack to write a program and implements stack using linked list using in C language and cplusplus lecture and saw all the concepts I have explained you in the earlier session on the whiteboard. Now in this session, this will be execution, writing program execution on the system that you can see how you can do different operations of the stack. Now, lecture three is to write a program for an executing for application on the stack call balanced parenthesis. So defensive and I have explained you on the whiteboard in the earlier session. So here we will write the program and execute on this. And I will check how the expression is having balanced parenthesis or not. Then in the same session, different lectures. Here is their program and execute in order to convert infix to postfix and to evaluate postfix and C language and C plus plus, there are different parts of it. We have already covered the concept and details on the whiteboard in our earlier session. This session, we'll be covering on the system. We will be doing hands-on on the system and write the program and executing on this system to achieve this application on this tank that is in fixed reports, fixed and motion debt evaluation of postfix, different parts, part one and part two are the same. Well, same application. Now, Lecture six is write a program in execution or in fixed. Now this was one in fixed reports fixed, this is in fixture prefix and evaluation of prefix in CNC plasma, again, we'll write a program execute on the system in order to convert in Pix2Pix are fixed and in order to evaluate the prefix using these languages, C and C plus plus, then. Now the new session will be for the different data structure that is Q. So yes, I should have studied all about in detail about the stack. Now this session we'll start for the cubes. Use theoretical and practical session. On the left-hand side, you'll see the programming execute on the system in different lectures, different programs are there or the view that is writing the program executing our queue using an array in C and C plus plus, we will implement queue using Eddy in C and C plus plus then lecture do is for writing the program and executing implement code to implement a queue using linked list in C language and C plus plus. Then lecture three is to implement the queue to write a program and execute for implementing you using some cooler linked list. So we'll see what a circular linked list, how to implement queue using the circular linked list. So these all are different, different programs for the use and these are very important and these are also asked in your interview exam. Different, different programs for the queue. So here I've explained you have executed in the system, you'll understand in more detail. Then lecture board is to write a program and execute code implementing circular queue using an array in C and C plus for the different parts, part one and part two. In order to implement circular queue using an array. Then Lecture six is for writing program and executing the DQ. You'll see also what is DQ and we'll implement it a culinary. See what a circular array then what is v2 and how to implement the queue using circular area in C language and cplusplus. Then in lecture seven, there will be proved writing program and execute, detailed execution of presidency or his gratitude dq, then you'll implement PriorityQueue using the linked list in C and C plus plus programming execute on the system. In this manner we will be completing. And then we will be coming to the different data structure that is linked list, linked list theoretical and practical session on whiteboards or this mean I understand on the whiteboard. In lecture one, we will be seeing what is their different, really see detail. What is linked list? There are different types of linked lists. The linked list, singly linked list is one other type of the linked list means see what is the singly-linked list and we will see in detail about it in this introduction in lecture one. Lecture two, and lecture three is your main function I have covered so far. We know that in C language and C plus, plus the entry of an execution of a program starts from the main function. When you execute your program, the control comes to the main function and nine by nine, the instructions will be executed. So the details about it, I have explained you in this lecture for the singly-linked list program. What is, what we'll have in the main function? Part one, part of the program. I explained you then in lecture four actually we will see there are different operations for the singly-linked list. So one of the operation is to add a node at the beginning of the singly-linked list that we'll see in lecture four. Then another operation of singly-linked list is to add a node, a new node after a given node that we will see how to add a new node after a given node. Then this lecture Six Sigma program for adding reform, that is to add new node before given node, how to add it? That operation we will be seeing in this lecture, six different parts product one, lecture sentence covered the part two of that program. Then lecture eight is what adding the given node at position, position will be also given. Node will be also given which node you have to add in this and how to achieve in C, the program, this intellectual aid. Then we intellectual nine, we will be seeing the program for creating this thing, how to create a singly-linked list that programming is missing in the lecture nine. Then we will be seeing how to display the singly linked list, how to display the nodes of this in getting his foster father Indian production really understand singly linked list. And what are notes that are notes in the same in English that all details. I have got one Indonesian lecture, how to display all the nodes of the signal in English. How good is failing singly-linked list deviancy if you see the program and we've seen this lecture ten. In lecture 11, we will see how to count the nodes on the linked list. How many nodes sound energy will see that programming will be seen for the same mean increase in lecture 11. Then Lecture 12, or displaying and counting nodes on the list will display the program how to display the nodes something is and how to count how many home, how to count the nodes on the list, singly-linked list. Admin seen the lecture 12, then, then different. Then we've seen the different lecture is there to cover a singly linked list program for deleting nodes. How to delete a given node on the singly-linked list that is in this lecture, then different lecture part for the program for searching your procedure. Such the node on the singly-linked list every receive the program for it in this lecture, then different lecture, we will see that a singly-linked list program for reversing how to reverse a singly-linked list that we've seen in this lecture. Then we will becoming to the WE linkages. So in the earlier lectures we were seeing about signaling is now different. Another type of linked list, WE LinkedList detail about it in this lecture, 16. In this lecture, that is a willingness in production. And another lecture will be writing the program. So there are different operations on the doubly-linked list then that we will be doing now in different, different lectures. This lecture is for doubly-linked list programming order to earned abroad venue or list is empty. Then when you have to add up new node in the doubly linked list, this is a program are the same. Then another program is to doubly-linked list program to add the node at the beginning. So when you have to know, you have to add a node at the beginning of the doubly-linked list. How to achieve that program is there in this mixture. Then another lecture is the doubly-linked list program to add at the end on the envelope. So whenever we have to add a new node on the, on the doubly-linked list. Then how to add at the end? How to add that program will be there in this lecture. Then doubly-linked list programmed to add. The mode on the listed is whenever you have to add a new node after a given node in the doubly-linked lists. How to achieve that program will be there here in this lecture that I'm explaining your new data. And then doubly-linked list program to add the corn or whenever you have to add a new node before a given node on the doubly-linked list, the program will be there. Then we will be seeing how to create a doubly-linked list and this mixture in this different lecture that you'll see the program for it. Then another lecture is to delete the nodes. Whenever we have to delete any node on the doubly-linked list how to achieve and see the program here in this lecture. Another lecture is to reverse the doubly-linked list. How to reverse a meaningless, definitely seen this lecture. Then, doubly-linked list program to how to display the DAP notes on the doubly-linked list that we'll see the program in this lecture 25 in the next session of this course. Maybe it'd be also seen the application of use that all the queues are used efficiently. So I'll be covering the lecture, one of the lecture for these applications. Then in next session may be seeing also the application of linked list, where all the linked list can be used efficiently. So also I have covered the practical session for these applications of the cube In my last link that is thought to add mode on the application of linked list. I have already important application is to manipulate the polynomial expression using a linked list. So I'm explain the theoretical part on the white board as well as I have also followed the practical section on the old block ID on you will come across this important application. And also I will be sharing with you the code for all of these programs for these data structures and application so that you can access it, you can refer it, and you can also implement it by the frame. In this manner, we will not complete course. We will be deeply studying understanding these data structures. I'm surprised as well as practically for this session I'm done and be ready for the Hertz sessions and lectures. We'll be going sequentially. So thank you friends. Thanks a lot. Let's meet in the next session of this course. Thank you. 2. Stack using Array C & C++ Session1: Hi and welcome to the new session that is stacked in data structures and unguarded them. So here in this session, given rewriting the program to implement a stack using EDI and C, as well as C plus plus on the code block IDE, as well as we will execute that ball. We know that this code block IDE, it is freely available. You need to download it and you can simply install it. It is very easy to install. And this is how you're going to get the IDE. And then you can pay the program and execute a proper hands-on and C as well as C plus plus. Now just before moving to the blog ID, Let's just quickly see the overview of this tax. So already we have seen in the session but a detail that is about the stack or the stack, how important it is. We have seen the real-world example. We have also seen the basic operations for the stack. We know that push and pop and always spoke bombed from the top. You know, that stack follows last-in, first-out, That is the last element which is pushed in the very first element. We know that we are having the top as a reference for, for forming Bush. And here we will be seeing all these. We will see, we will be writing the program to perform all these bush Bob. As well as we will be seeing the completion. This peak operation, we know that we will be returning the very last element of the stack. It will not pop anything. Then we will also display the elements of the stack. This is how we implement various functions. All the functions we've got acquired, implement the stack using the editor and we'll be running it. And C plus, plus as well as c. We know that when the stack is empty, in that case, the top minus one. So what will be the top in that case? And the stack is empty, it will be minus one. We know that in case on the edit first of all, the indexing starts from 012 till n minus one. Therefore, when the stack is empty, the heavy, we will be initializing this variable that has stopped to be minus one. And suppose now we have to push one element, that is the number. We have to push. In this fact, we know that first of all, all the elements will be on the same day that they've suppose we have the element of the integers for all the limits will be on datatype integer supposedly had proposed actor than all the elements of the stack will be on the data type character and so on. Depending upon the requirement, you have to push number five in my stack. In Italy, my stack is empty and I put up is equilibrium minus one. So in order to perform the push operation, and I am passing input number five. So what will happen in that case? In that case, you know that initially top is minus one. Because if it is empty, first off on incrementing by one, so that becomes 0 and that net position will be pushing my element. That is, I will be storing file in the subscript 0. Therefore, what will happen in that case? Subscript 0 because we know that the van I increment by one becomes 0. Subscript of that is area subscript 0. In that place, I will push, I will store the element that is five. So what will happen in that is their top becomes to be the men by one of the top will be equal to zeros. That is top will be roughly at 0th index and then this position. And you will be stored in the number of fights. So we will be stored in here at this position. So on. So now next time actually I have to simply push another number, say six. Again, my top will be incremented by one. So then we will store number six. This position, this Eddy. This is how we'll actually, we will be performing the push operation. So what about the pop operation to be not that we don't have any option in case of Bob if there are 56 elements stored in the array in the stack. So we don't have any, we don't have oxygen to form the value which was poached very first. Remember the pop operation is performed, it will be pumped, it will be posted, popping the last element, and accordingly, so we know that the stack follows last-in-first-out. It is the one which is pushed very last will be bombed very fast. We have to pump very first element. That is here. So we have to perform, in that case what will happen actually, we have coupon above that much amount of time. The 0th element. So you can understand the 0th element in the fog, the very large and it was the one which was pushed very first, but it will be found very last. Network stack is called as last-in-first-out. It obeys last-in, first-out on forecasted Pascal. So this is what happens in case of the stack for Chen. Also for the fall, what we will do, we will first off, I'll retrieve United. It is referring to the last element. So that value will be, will be, will be retrieving the top one. Even return the value which we have written. Some video of him. This is how we perform the pop operation. So let us start and directly write the program and then we will understand it better. So this is the program I have created a new project and the code block ID. I paused, I will be running. I will be explaining your line-by-line instructions in C plus plus, and then we will be executing it. And then it will be also seen the program written in the C language. First of all, there is no much difference and C and C plus plus because one difference I have used here just blueprint as to display as well as to input the functions which are called for C and C plus plus are different. Rest all the things, all the implementation, our scene. You will understand once you see the program. So this is the main function. This is the C plus plus file, in which this is a main function. We know that whenever we execute the program entry point is one. It doesn't mean function. Line-by-line instructions will be executed sequentially, which is dead inside this main function. So first-off on what I'm doing is I'm declaring these variables. The name you can see option and item. I'll explain you what is the purpose on them once we use them. Now, this is a simply display message stack using arrays, since we'll be implementing stack using Eddy in C plus plus. And this will, let me explain you what we will be doing first in the program. Since we know that we will have to perform this bookshop lesion pop operation, display, exit, all these operations we have to perform for the stack using EDI. So in that case, usual users should have the flexibility to choose the option. So we should ask the user these adoptions. That is, if the user press one and you don't want to poke on push our patient if he wanted to perform pop user to enter two. If you don't have to hop on. In that case, you don't have to enter at least so all these things we will mention to the user, so that user will be flexible. User can do anything which Bob IQ, and if you don't have to exit, we will provide that option also. So simply we will be using a switch case. And I didn't get an ink, the switch is in my one. This, you can understand. This loop will be running infinite number of time. The user don't mention to exit that option also have provided because whenever I am writing while one that means a loop is going on, it is running in finite time. So you have to ask you, is that actually the user have to come out of the loop, don't want to perform any operations. And the user is supposed to press any number. And accordingly, you will create the case for it since we'll be using which case, it will be getting the matching case for that. And we will be writing, we will be writing a few lines of instructions to exit from the loop, to come out of the program. Do exit from the program. So this is how we will be doing enough program. This is my L1 and you can see Environment quarters there. All. I'm having my switch case here. In this switch case actually better. And you can understand that it said that I'm displaying different messages to the user that Warhol options are then depending upon that user, have to enter that value. These values I bought actually since you know that in case of the switch case, what we do, It's a purpose or switch case. The purposes that first of all, if the user is having, have two 2's from n number of cases actually have to perform a number of operations. So depending upon what users want, we can simply create the cases for that much time. So having five options he was at evidence is having five options. So we'll create five different cases for that. Accordingly, we will ask user to enter that number. So you can see here that this while-loop, I've just displayed this message that is the smart, use it as having this option and user is supposed to enter these numbers for performing corresponding operation. And then this option actually that is which I will pass in switch. This is option that is 12345. That is the option. And you can see this is the medieval which are declared earlier that I will make US feared. In this switch case. You can see in the switch I'm making the use of this option that can be 12345, whatever the user provided. And then accordingly, I have created the cases, case one, case two, case five. And finally, we know that in such case, the default is that if it does not matter, then it will go to the default case. This is what happens in case of switch case. Let us see one by one, which operation for peak display an exit. Now, what will happen if we have poor performance and user one element? So you can see, first of all, I had in this program, I have first of all, creating the stack is a stack using the EDI. So I will be creating the eddy, that will be the stack. So yes, this is my start off my program. Since it's a C plus, plus some project, I have to, I have included this header file I O stream. This is what CNC out I will call these functions we not seen as stool input from the console, cout display on the console, input, output operations functions I'll call. So I have to include this header file, then this namespace, std, then whatever function we can use returns under this namespace. And then you can see Stack using edit that I have created this tack. In my case, I want to store all the elements of the stack on the data type integer that created the stack. You can see it's a simple array. Since this program and stack using EDI. That is the reason I'm creating the EDI. And this is depending upon your requirement. If you have to store in this stack all elements as the datatype breakfast. So instead of int, you can give it, as you can see here. I have given you up. If you want to store character, you can get scared stack, I'm just good at it. And this max, so max, which is that actually I am hash defining some value. So here I will just mentioned hash. Define max. Suppose Cathy, you can define to any value depending upon how much size you want in case of stack. In case of the ADA, you know that whenever we are implementing the stack using the eddy, we have to define the size homeless size you want. What will be the size of the stack? You have to mention it initially here. You can see, because you don't have any option to change the size whenever you are creating and implementing the stack using EDI. That is, whenever you are implementing the stack statically, statically we cannot change the size if you are using the eddy. What if you have two teams? In what, what? If you want the size to be flexible? In that case, you have to implement the stack using linked list. Since I'm implementing stack using EDI, the sizes fixed, I cannot change it that at anytime actually. As I told you for the push operation and the stack, the school and we are performing the push operation. What will be the scenario in that case? We have two Falstaff while check whether the stack is full or not in the stack is full, we will be simply giving the message. We cannot perform the push operation. And another nice that you can simply add one more eddy that the size will be the double the size of the old area. And you can copy the content from the old array to the new Eddie, and then you can pop on boxer. In fact, you can see how much time consuming it is, how much memory it will be taken. Here I post-op, while we will not go into that a manner, we will simply display the message stack is full so you cannot perform the push operation. So the best actors using a linked list, because in that case it is irrespective of the size, you can simply you don't need to bother actually about the size. You can. You can run time on site. Who can runtime? This is the advantage of stack using the linked list. Now coming back. Since I've created here, you can see here by the name stack on this Good Eddie. So we should give them meaningful name so that you'll understand what you are doing. What is the purpose? This maximum, this, I have defined this isotope on the size of the stack to be Cauchy, so I cannot exceed that size. I have to maintain the size for forming which involve I always have to ensure that it does not cross the limit. Let's see. Now, let's come to the main function. You can see here this is, as I told you, this ad is so here we will be storing. This is just tack. That is the area in here. We will be performing the Pucelle box, so we'll be storing elements in this Eddy in this stack. Now, the main function, yes. So suppose the user wants to perform the push operation. User is giving, is SCOM. Just entering one. What will happen in that case? Since this, these 12345 all list this option, all these values and we will accept an option. And accordingly, the sweet time passing this option in this Mitch and accordingly the cases are there. Therefore, the user enter one. This condition will be satisfied, right? You can see here it is giving since this is a push operation, so therefore we lost you. So first of all, what element the user is supposed to be wants to push. So therefore, I'll ask user enter element to be pushed. In this time. The user will understand and then you will give that element and that I'll be accepting and call this. I have created the new project in the code block IDE for C plus plus. This is the program which is written in the C Plus Plus language. You can see here we are having. Here I've just increase the font size so that you will understand it more better. So we know that whenever we execute the program, the control comes the very first function, that is the main function. The main function is the very first function whenever you execute your program, as well as C plus plus mean as an entry points, that will be the very first function to be executed. Line-by-line instructions will be executed there inside the main function sequentially. Let's understand the program now. What I have done initially when I come starting of this CPP file, that is C plus plus file, I'm first of all, including this header file that is hash include iostream in C plus, plus. This iostream means since I took, I will be calling CNC output, inputting on the console for display on the console. Before I have to include this iostream input output stream, which will have their declaration of this scene and CEO function's input and output functions, which tells me call it, which I'll be neutral women to induce offered. That is the reason posts you have to decline that header file. I'm also using the names namespace, std, so that if I want, I can make use of it. And you can see here that this is a second. So you can see here I'm defining mass to tidy. So what is this max that I will tell you? First of all, since we are implementing, began implementing stack using the edit. So therefore we have to create a array. And you know that in case on the Eddy the size is fixed. We have to mention when we write the program initially we have to define the size of stats because runtime we cannot change the size of the stack because the size of the stack is, it is, it is not possible to change the size when your program, in particular tree that has only been on leave and you implement the stack using linked list, which since we are implementing the stack using the array, we have to define the size. And that will be fixed throughout your program and your program will be run. So that is a disadvantage that you are not having the flexible size if you are using the area the size is fixed. So you have to take care of it whenever you are performing, which we will see ahead. So therefore, since I'm creating a stack using Eddie, Eddie by the name stack underscored Eddie. And I since I have to store all the limits which are of the datatype integer. So we know that first of all, the elements, all the elements will be having the same data type. So since I have to store integer, therefore I have declared here integer if you want correctly, you can, instead of n, you can write it as depending upon your requirement. This size I'm giving you this size I'm already has defined it. You can change here and starting on the program if you want to change it, change to 40. But Runtime Window program executes, you cannot change. You don't have that option whenever you are implementing it using an editor. Now, these things I have done initially and you stop using it as minus one. We know that we have already seen earlier that when the stack is empty, first of all, we will be declaring and initializing the top will be minus one, indicating is empty. That is a thing. And therefore initially this thought which I have declared as on the datatype and we will be having the value to be minus one when I initialize it, indicating the stack is empty. Now, let's move on to the main function, because the control will come from here itself. Now this variables I have declared, you'll see that we have made use of it. This is simply a display message stack using EDI, then this mile loop. So what I'm doing in my program, let me just tell you post. So since we want the user to have these options, you can see here push operation for peak display, an exit. And I see all the details about it, how the functionality goes on, and what it will do, what is the purpose of each function and the lines of instructions. So don't worry about that. So we are displaying this message on the console. So what we will do, these options are there if the user wants to perform different operations which are there for our stack. And we add asking you, these are the options for this bush Bob. So what we will do, we will use us which case? First of all, we will switch case in this while one, meaning that this while loop will be iterating. And you don't ask actually, build a user, don't want to exit. A user is having all these options and music and play with these operations. Analyst user don't want to exit if the user wants to exit. That also functionality we have done. That is the reason we're using this myth case. We had asked, we are mentioning these are the different options, 12345, I named the variable name opportunity. You can accordingly simply enter these options depending upon what operation you want to pop off, even if you want to exit, you can enter five. And in the case we have implemented all these functionality which are dead bodies options, which that is a purpose This mitigates, might be running IT environment. So this loop will be going on iterating till the user don't want to exit. If the user wants to exit, then simply we have also provided that option and then it will then relax it. And we have implemented all these operations. So in the cases accordingly. So let's see, that is a reason I am running this while one inside this middleman, that is a switch case. These adoptions supposed to be are displayed on the console so that users will understand what you have to enter to achieve Bob operation accordingly, these options which are there, you can see I'm simply using C and C and is used to except the input from the user on the console. So this is the variable which I have declared here. So this is the media declared here and making us off with whatever user enter, whether it is a one on whether 2345 numbers between thrombin to find user will enter and that will be elected variable, that is option. And accordingly, I will pass this, switch, that option and I have needed the cases 412345, whatever is there I created for that. You can see here case one, case two, case three, case four, and k is fine. And if the user don't enter any of these options by default, you know that in switch case we have this option and it will go here saying that user has entered the wrong option. Let's see, post for the push operation. If the user wants to perform push, user will simply interval. Therefore putting this which case will act, submit this case one will be satisfied because he was I'd have entered one. And therefore you will have to poke on push operation. First of all, you as opposed to ask the user to enter the elements that users have to foot. Therefore, you are asking you, is there that right here you can see an element to be pushed in this stack. So user will understand, user will enter that element and you are calling this suggest I have modified this actually, which I have written it as scanf. I modified it to scene because we are using, we are implementing the code in C plus plus. So therefore for C plus plus b, you'll see in NCL, seen as used to accept the input on the console from the user. If it was the C program, then we would have used, not seen actually, you would have used that scan f, which I've used allele giving the farmers as a fire like this. This is a V. If you want to give input on the console, you use scanf. You mentioned the formulas specific essence, we want integer, the item to be inserted, ability to type in the full performance. This is format specifier, network poison the different bombers specifier are there for different term datatype and this item and we have to pass and present I can You have to pass in this manner. Instance, I'm not writing the code is for C plus plus. Therefore, this is the function for that. That is C in C and is there to ask the user to enter, the inputs are intact. We have to not mention the hormone specifier directly. We'll give this variable name that is item. That item the user enters, we're collecting an item. So suppose the user enter five, so that would be a parsing. You can see I'm calling the function and passing this item, that is fine. So therefore the control goes to the function. Let us to the definition of the function and this value. Let us see. What will be the definition of the push function? So we will move to the definition. The control comes here. You can see here push, this value will be passed to this function as an input argument. Here. This will be very human. So you can see here this five is pass and this is on the datatype. And since we want to simply push onto the stack, we don't want to return that would return. You can see here this is a void. In order turning anything we have to simply pushing this time. I have already mentioned when it might be at pitching in the stack, we have to always take care about the size of the stack. We have to always make sure whether your stack is empty. You have to make sure and then your stack is not cool. If the stack is not full and only you can perform the operation, then only you can call the function, then only you can watch any element if your stack is not quick. If you don't check this condition, you can understand the status Fool and you're still pushing, you're not checking that condition, then the old law can occur and it can behave half as a result of a white guy. You have to always give this condition. So what is this? I'm calling one more function. This is by the name stack full so it fits in paycheck. My stack is full or not. It will return accordingly. So what should be the definition mean old and will be the stack. What we have, we know that the top is equal, equal to max minus one. That is my nobody who are top, as you can see here, suppose I am having, this is a stack of this ice. I'm doing hash define match tool six elements, and you can see indexing will be 0. Suppose my stack is food. Despite that are all that I have stored for five more elements that I have stored 678910. Therefore, my topic will be fine. When I'm calling the push function, I'm passing some value, say Bush, and I'm passing ten value in that stack. What will happen, my talk in that cases. And still I'm not making I'm not the charting whether the status will in that case, the older flow will occur because there is no size I'm eliminate will have half as many saw that we have Paul this checkered actually. And that is a reason. Then the stack will be called whenever my talk will be equal, equal two. You can see here this iss what I have given here. If there are six hash, define max six actually therefore I've created assess stack of size six. Therefore, since in an array, indexing starts from 0 to n minus one, therefore it will be from 0 to five. You can see here therefore, if my thought is equal, equal to max minus one, so that was six minus one becomes fine because the Internet indexing starts from 0. Therefore, I have to always be in short if my opposite equal to max minus one. In that case. It means that stack is fully in this function itself. What will be the definition of it? See, this is this I have defined here. It will return the integer whether yes or no, the stack is full or not. Dancer will be rather one on it will be 0. Therefore here I'm checking if OP equal equilibrium x minus one, then simply return one else and return 0. So I hope you understood what I am explaining you. Therefore, coming back to this operation. Yes. Actually, if my stack is not put, in that case, what will happen? Definition that I have implemented this. It will return 0. This stack is not full and this condition would not be satisfying. That would be weekend push any element because the stack is not put. In that case what I told you what we're supposed to do first, we are supposed to post increment the top by one. We have seen that. And then we will add that base actually then, which the user Wanted to push. And therefore, you can see here, since we haven't created the EDI on the datatype in by the name stack on this area. And more to stop actually, initially top is minus one. But now at this line when dog gets incremented by one becomes 0. Network stack, stack on Discord add a subscript 0 is equal to item that is despite that would add 0th index. Actually this guy will be stored. Therefore at 0th index. What will be stored? This five will be stored here. Because stop is incremented from minus one, incremented by one. The top is 0 and liquid at this place I'm storing this element. This is what we are doing in case of the population, and then the stack is full. This condition will be satisfied if the stack is full before pushing this candy. Before performing these operations in the stack is full, then this condition will be satisfied. So again, if I find poaching, if I'm already put all the elements, my stack is full. Now again, I'm pushing some more elements, then it will come here. This condition will be changed. Yes, it is full. In that case, it will enter inside this block, it will give me Stack Overflow and through the turn from non perform this operation. So this is only in the case when the staff. Is this condition we have seen in this function. We are checking if topic will equal to max minus one and only this full, this is the simple push operation. Let's move on to the case for the push vet in me trying to understand. So this is an then I have given this big, you know, that in switch is for each cases we're supposed to give the break in this manner. Because if you don't give, the next case will be executed. So you don't want that because there are different, different functionality for different cases. If the user wants to perform push, only push should be perform. You don't want you when you've been not drive until and unless user don't enter that option, you have to mention the break. Then for the pop operation, let us see now what next for the pop operation. So if the user wants to perform pop yours, user will enter to switch. In option there will be two. So in this case, two will be this fight. And therefore, we are calling only the par function. You can see I'm not passing any arguments since we don't have any option. The last element will be on Nipah pt. Depending if you have to pop all the elements you have to call that much amount of, you have to call the pop function. That most amount of time. You remember, just last element will be formed. That what I'm calling pop, the element will, which will be bought. So let us see the definition. So what I'm collecting in this video, but this is item which I have declared as the datatype int I'm collecting here in this spot. You can see here. Yes. So let us see the definition, what it is doing and how it is returning Information, Bomb devalue. Let's, let us go to the definition of the pop function. So it doesn't build definition. So again, One more thing. Whenever we are copying, we have to answer check whether the stack is empty. If your stack is empty, in that case the top will be equal to minus one, right? No elements present in the stack. You obviously you are not supposed to fall because there are no elements. So you have to check before before writing the instructions to perform the job. You have to, first of all check whether your stack is empty or not. I have tagged by the stack is empty or not. Let's see the definition of it. It is what will be the condition we have to check? The stack will be empty. We know that this is in addition, if topic will equal to minus one, that means the stack is empty, so it will return one, it will return true if the stack is not empty. Therefore, in that case your popularly not minus one, it will be having some value like 0123. If there are some elements present. Net case it will return 0. Now moving to the fall. So therefore we are checking this condition. If this condition is not satisfied, that means the stack is not empty. There are some values present in it. Now you can fall. So therefore, this condition, this will not be executed because this condition is not satisfied in that case. Therefore, what will happen? You are simply this instruction will be executed. What is this? If we have football, first of all, we have to retrieve the value in some video bill, and then we have to decrement the top one. So simply I have made it in my online. That is what I've done. I'm simply using this area. You can see n subscript minus, minus. Minus, minus means, it means you, I'm doing the post-implementation, that means fast. It will be acting like this, like stack underscored, add a subscript. In the next line. It will be topic while to talk minus one. That means we know the basic thing whenever it is opposed agreement that variable will be making use. Variable is first used and later variable will be decremented. So therefore, what will happen? I'm returning this element that Is that better their toxicity frame. So the top will be referring to the last element. So that will be, I am returning that value. So suppose that only I have pushed five on me. Now whenever this stack on this Good Eddie subscript, only one element on the app which you know that in that case they're talking Vizio stack underscored at a subscript 0. Subscript 0 is what actually, you know, that the top or top is actually 0. So what will happen? It will return the element that is five and then I'm decrementing by one because it's supposed implementation. So top is 0, count becomes minus one. Again, see, instead of this, you can also write like this. Item, like this. And simply item is equal to stack underscored. Edit subscript. Simply minus margin. Simply return. You can also write in this manner. You can see I'm declaring an item, whatever the stack underscored. Subscript office stop is what actually suppose you had only one element is present in the stack. So in desktop is you donate to stack underscored, add a subscript 0. Suppose that element is only five, which is present. So therefore it will be stuck and it's good at is absolute 0 will be giving me the answer. That's fine. That either I'm collecting in the variable item and then I'm doing talk minus minus. So Tom becomes from 0 to it becomes minus one. When I decrement by one, then it becomes minus one and I'm returning this fight. This is the thing which I had converted into. Second one line. You can do simply in this one line itself. Instead of all these things. Again, choose whatever you want to do depending upon. Now, this is how we poke on the ball. Now moving to the next case and now let us see. Yes, now it has return me some. And you can see here, the bank is returning this element that is fine. Datatype is in. Now what will happen? Therefore, this value is that element is returned, is five, say Nike's. Therefore, we are collecting in the variable item and we have seen a dissolve the datatype integer. And we asked what we are doing here. We are simply displaying the message bar value and that item name. That is the variable name that is item. So therefore that element is what files. So you will get simply BOD value is fine. You will get the element name, element which you are copying just to display so that he was OK and make sure yes, this is a correct value we just bought. It will be the last element which will be false. If you want. This is how, depending upon the requirement, if the user has English five elements in the stack, user will enter one and then we'll put some element. And again, user had to perform push operation again, the user will enter one again with some element. Again, the user wanted this five times. The user have to push some element that will be pushed in this stack. And after that, whenever user one to pop, the last element which was pushed will be four, will be formed the very first. So this is how it works. So you can simply write the program in the same. I'm not in C plus plus N venue executed. We will see how flexibilities, how it will store, it will relay motif last element which is which will be formed very close and the very first element which is pushed will be pumped. Very last military this. Now next, let us see the next operation. That is, we have seen what is the purpose of the buses. It could return the very last element on the stack. It did not pop. It will do the same operation what Bob does, but it will not pop the element. It will simply return the last element. It will not decrement your top whenever you have seen in the fall and he was returning the last element. But we must also decrementing the top in case on the peak, maybe simply return the last element. You will not decrement dot dot. So therefore, here you can see what I'm in the case three will be satisfying. I will switch on, switch d will be passed case we will be matched. And we simply displaying last element in the stack is the speed function. We're calling this function because this function simply return the last element which is present in the stack that only I want to print, and therefore I'm calling CL. And let us see what peak middle written obviously should return the element of the datatype integer because we're storing the stack with the elements of the datatype int. So it should return me some value. And therefore I'm displaying in this manner. So let us see the definition. What we have to write the instructions is z. So this is my peak function 3. Stack using Linked List C++ Session2: Hi, welcome to the new session. That is stack in data structures and algorithms. This session we've been writing the program to implement the stack using linked list and steep as plus realized as well as we will be also executing it on the woodblock ID. I will be also mentioning you the minor difference if you will be writing the same program and C language. Before moving to the code block IDE and raping the program, Let's understand it. Revise the basic things which are required prior to writing the program. We have already seen when m and we have implemented a stack using linked list. In foster one, we know that stack follows the principle that this last in, first out, that means the element which will be pushed lastly will be formed the very first. And we know that in case of this tag, the push and pop operation, it occurs at the same end. So here it will be at two. Since we are implementing using the linked list, we know that it is efficient to perform push and pop operation at the beginning of the linked list instead of harmony, which in bulk, and we have seen the problem if we perform which involved at the end of the linked list. And that is every time for each portion of the linear element we're programmers from starting with the end of the list and then we have to insert that element as well as for the pot. If we IPO for me, that is if you're performing. And then we need to also perform pop and the end. And that we have drivers. And then we have to delete the node that is delete the element that is pumped up illuminance, you can see there is a reason in case of the linked list, if you've seen it, we are avoiding to push and pop at the end of the linked list. Instead of pushing and popping, let us, we will be inserting and deleting the node at the beginning of the linked list. I hope so you have gone through the previous session of the linked list wherein we have seen that you have seen how to create the linked list, singly linked list and different types of meaningless. So here we're only making use of singly linked list. So how to insert a node at the beginning? How to lead in order to link these operations will be required. Also, I have given you in the earlier session that assignments to perform these basic operations. I hope you've gone through the linked list session and that will be very much helpful. If not, I recommend you to go to bed. So let me see here. Insert the mortar the beginning how to delete. So here in case since we are implementing the stack, the Insert means pushing the element and deleting the node means popping elements. We will be using all those instructions which are required in the linked list, all those operations. And we will be seeing it here. What happens in each allele, but stack is empty. That means since we are implementing in the linked list, we know that to implement the stack using linked list is implementing the stack dynamically. That means you don't need to bother about the size. You can decrease the size, you can increase the size runtime, you can allocate more memory, you can deallocate the memory. You don't need to bother about the size as we have seen when we implemented the stack using EDI that decides we have fixed. Before executing the program. You cannot change the size if you have already, you already in case of stack using any of the sites in your program and runtime, you cannot change the size. So that was one of the disadvantage when you implement the stack using letting or in case of linked list, the advantage is that you don't need to bother about the size and how you will be allocating the memory simply, you will be calling the malloc function in order to allocate the memory in the heap. And for the node, whenever you have to push any element, it will be simply first of all, allocating the memory for that node and then pushing the element. We know that in case on the linked list, the nodes are present. In each node will have two fields. Boastful will have the data, and the data type that is np TO characters. But we know that all the limits on the nodes will have the same datatype. It is either if you are filling all the modes in V2 and then it will be integrated fulfilling all with the corrected will be correct. But the next field of the normal appointed with the next node so that we have seen all these basic things. We have seen. You have a word in the linked list session. Now, consider your stack is empty in this manner. And in that case we know that pollen in linked list, whenever the linked list is empty, we are representing a start pointer that will be a high head point, then it will be also not. But when we start creating the very first node, in that case, we look at the memory and that stock will be pointing to that portion of each and every time. We have to keep into account that the staff will be always pointing to the first node of linked list. Here the start in case, since we're implementing using stack, we have to always refer by the name topic of staff is a reference point wherein we will be pushing and popping me know, in case of the stack. Though, stack is empty, that means there are no nodes which are present. Because we have created the Lao and Blackboard will be null. We represent instead of Scott, you're using the name has stopped since we're implementing the stack will be null since the stack is empty. Now, what will happen next? Whenever the user will call the push function, that is in this manner. The succinctly. Suppose you are having this calling function, and here we will be pushing the element of the datatype integer. So suppose a user is boiling this, use them on to perform push function and use it by seeing this element ten, user one actually which distend and this this stack. So what we have to do in that case. So in that case, first of all, we have to create unknown. Since we are implementing a stack using linked list, we have to create the node. Yes, we have to create a node and that is dynamically, that isn't the thing that we have to allocate the memory using manual function that will be the postdocs. You have to create a new node. And then what we will be doing, we will be simply, we know that it said that for speed will be having the data. So we'll fill the element with the data. That is, it will be here. Then we'll be filling it with ten. And what next we will be doing. The next will be simply point to the next node since it to somebody first node. Therefore, we know that it should be null in that case. Let me show you how does it looks. Initially our stack is empty, then focus on so when you create a new node, what changes will be happening? Let me see that now. This is the thing when we create a node, when we add a node, in that case, what will happen first of all, stack is a stack was empty naturally, but we have the user. The user wants to push the element. Then we have seen, so we are creating this new node dynamically allocating memory. So this is how it looks logically our stack. So this is the new node with them, given what you will do your fill that element one field, this is one node we know in ESOP linked list. So the first field will be paid with element which the user is passing, that is ten and the second field, since this is your first, note that with the second field, you're not having any other node. It will be, it will not have any at this node, it will be simply, you can see here. Now what you have to modify. Now you can see this is now a stack. Stack is not empty, but stacking stack, you have already created the node, you have finished element. So therefore, you know that the stack is reported by the thought should be pointing to the body. It should be pointing to the very last element which is present in the stack. We know the stack follows the last impulse down. And so therefore since the topmost null initially when the spec was empty now your stack is not post-Snowden. Bob should be pointing to this node that you have created. So in that case, what logic is required? We will also see in the program I just mentioned he has for your explanation purpose. So this is the launching. This is the definition on the push function matter if this is your new node. So I'm not written, not shown how to allocate the memory for the new node we will see in the program radically when we write it in look or block IND. This understanding we created a new node dynamically by using malloc function, and this is your new node. Once you allocate a memory, then there are two appeals of the node. The first field is beta and this is item at this span will be filling this ten. You're feeling this data with this value ten, which the user pass. The next field of the new node is what next? We will initially, since it's somebody first node. You can see here this is a forest node. And therefore you can simply, you know, the next and then next. The second field of this node should be null because it's still false. No, we don't have any I didn't not know. We just should point to and therefore, we know that it is a null. Therefore, we also know that top is also null initially men mistake was empty. You can simply write in this manner. That is the new node next is equal to the new node. Next is equal to five. That means simply, you know, negative office multicore, this new node next. That is, what is the next? If it's only actually it will be null itself. Now, the next thing, what you are, what you need to modify. Now since you have added one node in the stack after the talk Chuck point to that node. So this is a new thing which you have to do. Your next line of orbits you ever do that torque is equal to new node. Here. Why we are doing all these things? We know that now the talk will be since now the stack is not empty. You are having this always known presence. So that should point to this new node. So in that case, what will happen? This will not be null. So it will not be equal. It will be not null actually. And you will be having simply the top. Here. This is your top obvious now pointing to the new node. So you have given in this top is equal to new notes so that this new node which you have created and not simply write equivalent new node. You can see Dr. B. Pointing to the very first node, which is what we have to also ensure that we have to do this pushpin. Recall the definition of the push will be in them and I add a node at the beginning of the linked list next time. So you can see here not only one node is being added in your room stack that is, one element is pushed in your stack. Next time when you render user again call the push function and use a need to push, say the value 20. That is what you have to do in the implementation on which function. We know that since in our linked list we have to add a node when we call a V&V afterload would approach and then we have to simply add an order depending of the linked list. So therefore the, it did you see the next time the user, also called the which function and user needs to push this 20. Warfarin happened. So obviously this 20, We have to push here actually at this place, we have again create the memory new node. And then we have proof in one field, first field of the node with that 20 and what will be the second p. So what we have to take care for that? We know that since we have to add a node each time, let me create a new node. And that node in this stack. In that case, we have to add these nodes. I'm thinking of the linked list network. If you represent this node, this one, this was the president. This is your logical stack if you simply, I have also represented in a horizontal manner. So this is your vertical manner because each incus, since we know that in case of stack, it will be represented in this manner. But we have seen in the linked list, this is your new node, which initially when you added this force node and next time when you add 20. So we know that since when we push the element 20, we have to perform the operation to add a node at the beginning. So this is your personal menu. I wanted to add the node at the beginning. Then what should it be? The case like this? It should be In the next node, which just a second. Therefore, you have men again, he was called up which function and pass 20. So this will be added when you represent these pink tactility in horizontal manner, as in case of the linked list than this 20, which is bad. This node which you create actually will add at the beginning of this in the linked list. Initially only disperse, notice present. Then you add your the egg, the next node actually that is a new node at the beginning, that is not yours. So you have to not after the second node, that is a new node. After this node you have to add at the beginning because we know that whenever, if you had calling the push function, we have to simply implement the functionality of adding the node and the opinion. So each time on DNI, I've been simply added the new. So this is the node which I'll create. And then this 20 actually, so this will be the new node in that case. So therefore the new node we will be adding beginning of this node actually, and therefore this will be my new node. I'll create a memory upon the new node and I will fill before spilled on this node with which the user is providing that is 20. And what will be the link part of this node? That is, we are referring it as by the name S. Next, it will be simply the address of this node. You can see that again when I create an user again, say Part D. Again, we'll create a new node. And again, we're going to fill it with the force field with the data which user is providing. And the next field will be the pointer to this new node. Because since we need to make sure that whenever we are pushing the elements, we have to simply poke or in case of the linked list, we have to add a node at the beginning. So we have to not know each time the node will be added at this place. That is this. Now you can see here, since we are, when we are users pushing 20, we're simply creating the node and this node at this node. Then we represent, in that case what will happen. Therefore, the new node, that is this node, this it's linked part, will be simply having address of this node. That is simply, you can see that new node link will be equal to. So what is the address of this dog? Because 100, So that whole new node link will be equal to talk. This one. You can see your new node. Next, we are representing the by the name is next part of the node and it will be equal to top. You can see here when we do in this manner, then what will be happening? This will be the thing that is a new node. Second part is the link button. There is a next we will have of this node actually that has stopped. So therefore it will have 100. And you can see that new node link is pointing. Now you again, you have to want to change. You have to do sophomores. Pointing to this node, captured point. We know that we have to make sure that top point to the new node. That is the last node which has been added. So the last node is this. He said, Notice this one. So therefore, what will be the instruction? The same thing, that is this one. The top is equal to new node, not be pointing. This will be this will be the top on this one. This we have done. Again back to me then we have seen that against how it in this target will look, look in this manner. So in this stack it will look like this. This is joint D is Edit. And you can see here this is the horizontally form. So the same thing, that new node which you have added actually, and we have seen that a new net, new node next Fed is linked bark will have the address of this node which we have, which was computed earlier. And then this will be updated. Ofs is equal to new node and depo not binding to this mode. So you can see that we are making sure that every time whatever node we add will be pointing to that node B and making sure that linked list, the node seed, the standard was already false. Then the next time we edit 20, so we are adding at the beginning, that is before this, this node, we are adding this new node, which we are creating the ID before this node and not after this node. So that is we're not an ID at the beginning of the linked list so that we have not private. We're not, we're maintaining the adhesion always at the beginning so that we can meet weekend and China, we achieve a good diet that is very much efficient. That is, the time complexity should be big. One. This is also a next time. If we want the users simply calling the push function. And again, he was user again, foster to user one to simply, again and again push this cart in the stack. What will happen in this case? Again, you know that what will happen again, we will create a new node here. Again, the limit will be located. I'm on that new node will be filled with for the next belonged new norm, it should be pointing to this node. Therefore, this will be the condition that is a new node. Next is equilibrium. Request. Top is pointing to this. Let us know earlier node and then the new node we have created. Now the top should be pointing protecting mode. So therefore torque is equal. Newnode and NVD present horizontally. That is, we're simply adding making job Here. We are. This is just a representation to make you and explain more better. So within the stack, it looks like this venue. Venue is simply represent in the horizontal form, it will look that you are adding the thing that you always at the beginning. This part, it will be added before this node, this new it, an order will be added here and you'll be modifying that link will be the address of this node. And that is the next part of that. And then stop. I'm pointing to that node that is, whereas I told you, this is how this works and how you perform the pop operation. In case of the pop operation, just a second. It will be known that whenever we are calling the function, it will return the element which is being formed. In that case, we have to simply perform operation of the linked list that is deleting the node and at the beginning of the linked list bleeding the node at the beginning of the linked list. So therefore, we know that we are always having the top pointing to the last element in the stack. So therefore we are having a pointer. Is pointing always. Buddha, recent node which was been added so that whenever we are popping that not only will be what will be the change in water will read the instructions in that case. So suppose we are calling the pump function and we have used two elements, we have pushed into the stack. So when we call the pop function, we know that this will be reported where the top is. Top is adapting to the last element, that is the recent element which was being added. Therefore, this should, this node should be delete it. So in that case what will happen? We have to also make sure that we free this basement me, delete this. So in that case, when you delete this node placed upon, you should also be it sold. The stock. Actually, when this node is deleted, should then be pointing to this node. Let us know what it is having them on US. Debt preferred stock one, since you have to delete this node, you have to keep the reference software. What we will do, what changes, what code is required? Simply the code is that what we will do. We will damp node and we're going to be simply third focus pointing to this node, right? So we will keep the backup of the storm. That is TMP. So I'm just explaining in short the program even see how NADH has been initialized alpha-keto. Remedy than anybody I looked at the memory. So here simply big, big in the backup of the stop with this point though that this M by M taking this backup because I will be freeing it later. Now, the top should be pointing to this node. So what I will do is simply we know that Bob should be pointing to this node when we delete this node, which is having twenties should adopt to be pointing to the normal it is having to spin up. The next part of this nerve thought that it stopped early. Is that a stock? Next, it will be simply we are having the address of this node. How do we reach to this node? This manner that is the thought, thought me like this. In this, in this manner, the top and not pointing to this node, it will be pointing to this node that is ten n. What we will do. Finally, we will call the function and we will be free this EMB, which is that. Therefore, since we more business BNP, we wanted to having this backup of this node. So this via freeing the memory, since we have to remove this link, this is the instructions which is required in order to perform the functions. So I hope so You bought then this node will be deleted. This will be deleted, so it'll be left with don't need the SA node, which is having the value as ten. So I will move into the code block and then sinking with what we have understood till now in our diagram. As well as the concept services, my code block ID. But I have created the project or the stack using the linked list. And we'll be writing that. We'll see the program in C plus plus. I'll explain you instructions line-by-line and we will be executing it and see its output will see the basic, all the operations which we are performing butcher shop and what volt is required to implement the stack using that LinkedList. At the starting of the program. This is your main.cpp. This header file we have been through our program, we have written in C plus plus. So first of all, we have to include this header file, iostream because we will be using the function of input and output in order to simply inputted on the console. And two output on the console that is in NCR will be using. So that is the reason I had mentioned this header file. And we have to use this namespace, std. Then we're simply creating a structure since we are implementing the stack using LinkedLists, we know that in linked list is represented by nodes. And notice having to data fields present in that is supposed to be the data. And the next one will be a pointer to the next node. And therefore we are taking struct node pointer. And this can be anything if you want to be the more Mitchell Hamline guys character instead of and you can give it as an don't use, I will be putting the character, but here I'm taking it as NPV. I want my notes to be filled. We have the data of the datatype. This is a thing then are you can see here since we are implementing stack top is the main thing which is MRSA, evidence Madden, we'll push and pop. This is a reference point. So therefore, I'm using this top gustation maintained in the stack. Initially since the stack is empty, Nicola and a struct node on the top is equal label. Instead of, instead of this extra line in here also, I can write that pointer top is equal to null, but I have written in the separate line, depending upon how you have to write your code. Now, you can see here now coming to the main function, we have seen what kind of files that require yes. This is a main function. We know that whenever the program is executed, the control comes. The main, the main function. It says main function is an entry point to start an execution on the programs and pull all this comes in this main function and you'll program is executed, line-by-line instructions will be executed sequentially. So since we mourn different, different operations on the stack like IQ displays all the one who came all these options to the user. So therefore, what do we have to do? How we have to do the coding? We will simply use the switch case. And we will run a switch case. We run within my loop so that you can see this my loop. And here you can see I've used this switch case. If I stop on my arm using my loop and I've given this one because I am one minute to be born. It didn't repeat it into performed iteration and iteration till it should not end. The slopes are not. User, don't want to end it. So if you don't have to come born one to perform any operation, then you have to put them out on this loop. So I'll provide this option also off exerting so that user will exit. User will act accordingly and then use a mill exit from this loop. In that case, this option I'm providing to the user, user don't want to perform these operations. User can simply enter file. And in that case, we will be exiting from the program will be coming out of this while loop because we have to make sure we have forgive the stop condition to the user. So they're in order to not go to the infant. In order that user don't want to do anything, you do not want to perform any operations. This is why in one and this is both options which I haven't be displaying on the console. So I have c out simply. Here. You can see in the main function this declaration and we'll see how to make he was offered. But in the wild one, the message that he was a well-known that to perform push operation. User enter one, then user to enter two for 12, user to enter three for one user in before then display operation user have to perform. If you don't want to exit, you usually have to fight. And if he was at grid display, then user had before. So these are the options which should be up to Whiting and these options. And you can see a scene that is, we are collecting in this medieval option. This is, the policemen are making use of it. So we have declared this as in Teach up and these options, that is 12345, these are the different options that we are collecting. And we're passing in this switch, in this manner, we are asking the user is by using scene. By scene. The user will give the input and that will be collected in this option. And then we are calling in VR a passing in this switch, this option, that case one different. That is, since we are having these five options, that is 12345. So therefore, you can see that our five cases, case one case, two case, three case, wooden case. People aren't One. One, What is the one for push operation? If the user wants to perform push operation, then you usually want to poke on push. So therefore, in that case the user will enter one, and therefore the case one will be executed on switch and this display message will be there that enter the elements will be poached. So we are asking you to push water. Even user is supposed to approach that we use a one proportion the stack. So that is the reason I'm using Darcy in actually. And then I decided that I'm making use Opera. You can see here this is all the datatype and use them in any number and in item you installed it. And then I'm calling the push function and evaluate that pseudo support user is entering ten, so their quota, and I use a 12 and push their ten in this stack network I'm calling function and I'm passing this item let us spend. So let us see the definition on that. We have already seen one millimeter definition in the non-profit itself. What we will do here in case of push eyes, I have told you that we will be first of all creating a new node. And this new node we are creating dynamically. Therefore, you can see I'm using this malloc function that is inside this malloc function, the sizeof struct node because we want to create a memory that will be done. That is the head of the size of the stroke. No struck note is having Though fields, that is force field as data. Next few days, next bit is appointed to the next node. That question mentioned the size and you can see here, I'm typecasting struck, struck node pointer because malloc returns a void pointer. And therefore I'm typecasting it to the struct node pointer. So this is how we are allocating the memory for the new node. No, we had also checking that if you notice equal equal to none, I'm checking this can happen. This is good practice. Whenever you want to locating them in summary, always check if it is or not. That means in very rare cases it will give you not only men, there will be no space in the heap. In that case, only give you. That means there is an edit existing by creating the memory in the heap, then only this condition will be satisfied. But this is an exceptional case, but still I have to make my program very good. Should be edited checking for each, at each and every point. That is the reason I've given this is a good programming practice. You have to make sure whether it is null or not, because it can happen if there is no space and still you are using this new node and the question upper, right? This checklist meeting necessary, that is, if new node is equal to null, then it will display this message. Now, once the memory is allocated to this new node, then I'll fill this new node. That is, it is having uses data and next, so beta will be bored. It will be done. And even in which the US didn't have passed. So deaf wouldn't same thing and the next will be taught. We have already seen that in case of the poaching function, it's a second. Then we add pushing ten, then we stopped will be new node. We have seen that the stock cost of the new node next will be the top, should be pointing to that new node so that for this top is equal to new node. I hope so you want this logic you have already seen in our diagonal and so initially went and was pushed in bed this new weekly I did them remedy for this node that is envy-free ahead. And did this, we have simply filled about cost of yield on the node and the user. And third, and the second field. First node that we deserve, it should be dark because initially when the stack is empty, the top business depth adopt this new node that is now UI stack is not empty, this having only one node that span initially, and therefore they should be pointing to that and therefore gets it to opposite building new node. Same thing on the net is when the user is, I'm installing one to again push any other element. Therefore, user will give this value that is 20. Then what will happen in that case? This new node will be created. We'll create a new node, allocate, a new, allocate the memory for that by using malloc function means spin by 20. And the next field will be simply the address of this node. That is rather tough Was that, that is this instruction. It should be. Then you ignore next will be quieted down because we know that when we pushed this node red and then talk was pointing dead end when we have created the new node wherever we want 20 to be pushed. Then therefore in that case, the next next field that is shouldn't be pointing to the stop. The next instruction, that is, their will be new node. This node, the top will be updated to this new node and not destroyed. This is how we are performing the push operation. Let's see other operations which are present. Now. The user wants to pop bob dot node, in that case, what will happen? Then in that case? In that case simply the case two will be executed and we will call the function. And also you can see we are collecting in the item or this item it is operated by. Now you can see item it is on the data type in. Let us see the definition of the bulk function. You know, if I'd never pop is very recent and humans will be formed. You have to perform population in such Germano and deleting them. For first of all, it's a pump function and these other things which we are doing. We have seen that whenever we have to pop, we have to foster fall keep up on the top in some other pointer. Because we will be freeing that node, which is div at the top is pointing. And that is the reason you have to first keep the backup of their talk to some point. You'll notice the reason you can see here I'm taking one more pointer that is PMP and fascia holding before the popping Benjamin. First of all, I have to make sure that my stack is empty or not. Why? Because if the stack is empty, the stack is not having any elements. There is no chance to perform pump, there will not be executed. It should, we should not exit execute when the stack is empty because I didn't know and demons and there is no point to anything. You have to don't take care of this and you have to check this whether the stack is empty. Gps and you have simply exist from your program. Stack is not empty something some elements are present. That means you can perform the pop operation. And therefore you can see I have simply backup of the stop with the CMP. Mr. Pickin here. Here is pointing up this point. Now, what will happen? I've also. Element in some variable and after return from the function. So you can see here the stem for data stamp is pointing management office pointing its data. We're collecting in this value and this is the datatype integer. Then you can see later on I'm told I'm bringing this PNP. I'm feeling this DMP edit off was pointing because I have to simply delete this node. So if I'm calling for function, and so what I will do here, I could delete this node, is pointing to this node which is having the element Asteroid D. I delete this node. I have what I've done, I've got in some another point, I'm simply taking the backup of the stock to induct them. Then what I'm doing, I'm simply retrieve the value from M, and this is 2003. And then I'm simply free. Then top is equal to top. Next, that is in this manner, you can see the enthalpy is equal to talk. Next, we have studio, we added an impact. I hope so you understood also the pop operation. Now the next operation is not amino, peak operation. In peak operation, first of all, it will return the element which is the last element in the attack. Here first-off on the difference between he can focus that people only return the last element, but in case of the ball, it will return the last element as well as it will pop necklace tell immediately delete that node. But here peak will not do anything and just return the last time we'll display what mass element is present. This is OP cooperation. Let us see the operations. So that is three will be executed, function and medically calling and imprinting it because it will return the last element of the stack. So let us see. Here we're calling the function there to speak. Let us see the definition offer. Therefore, this is the definition. Just a second. This is my beak function. You can see here this one. Here. First of all, the return type is int, because we'll return the element. We are checking if the stack is empty because it's going to happen. There is no element in the stack. So in that case we cannot return because no element is present. Therefore, we have to make sharp before performing any operation. The stack is empty, then it likes it. If not empty, that means it will return a pointer to data. That means they're taught medical office. I'm pointing. You can see our top is pointing here. What does after return this 20? So we know that it's 20 is the first pillar of this node and it has to be present in, by the name as data, the next pillar of the node is represented by the name next what we want, we want only the parts with the code. You can simply returning data that is either type in, so therefore, it will be displaying it here. So this is what we are doing here. You can see you now in case of the case for that is if the user wanted to simply split, every user will enter or splay operation will be there. So how to display? So let me show here itself. So we know that logically in this deck it looks like this when you add and you replace it horizontally. You can see, suppose these two nodes are present. You notice top is pointing at this node. And whenever you have to display your whole linked list, that is which is there 4. Application of Checking Balanced Parenthesis C++ Session3: Hello, welcome to the new session. That is application of this stack in the *** structures and unguarded them. Here you will be writing the program for checking whether the given input expression is having the balanced parenthesis autonomic. So this is very important session because we will be writing the program are the same and we'll be executing the code block ID will be writing in C plus plus. This is the important application wherein you can use this stack of books solve this problem. Whether the given input expression, it is having the balanced bed emphasis using the stack. Before writing the program in the code block ID. Let's revise something important. Let's see the logic, what we should apply. I have just drawn here, thus stack, so amino not nearly the stack. We will be using stack using EDI vendor stack is empty. We know that the top heavy that thought in case of stack, and we know that in stack and pop operations pattern, the bush on the element will take place from the top, as well as the port will be also taking from them that the stack follows the principle last-in, first-out, that is there any minute pitches botched Verdi, lastly, will be bombed very fast. Since it is an EDI we'll be using you're adding, you know, that indexing starts from 0 to n minus one and all the elements will be on the same day that I'm answering. This has given expression. Suppose we are asked to find that, that is this expression which you can see here. Opening curly braces then by multiplication, then opening round record when people as foreclosing on record and closing curly braces, you are given with this input expression. And you went out to check that this input explanation, because having the belly button in on what does, does it mean? Balanced parenthesis. Parenthesis means you can see here opening curly braces, then opening round record, closing down, and record closing curly records. So all these are called as parenthesis. So we have to see that each of these parentheses, which is present in this input expression, whether they are having its corresponding here, this opening polymerases, whether it is having its got his morning closing. Yes, you can see at the end we're having this curly closing bend, as you can see here, we are having one of the non brackets. So whether it is having its corresponding clause, that is also called as right pattern applause, right? Parenthesis, yes, it is having the round Record, that is the closed round bracket or you can see it is having close parenthesis. This is what we have to check. So here this is balanced. If I put something like let me just open the door, this is the input expression. So here the same expression which I have taken here week within each of these parentheses article is opening his having the corresponding colleague Louis parents. So if you are given this expedition, say you are input expression is this one. Whether this particular input expression, you can check whether it is a balanced parenthesis or not. You can tell me it is not balanced because we are having here opening Brown record, we are not having, it's got a spondee closed round bracket, that is this opening. Left parenthesis is not having. It's got as funding round, right? At emphasis. As well as we can see here, we're having this curly braces close, but we are not having its matching open megacities and this is not imbalanced. But this is what about this expedition? If you are given with this, this expression is also not balanced parenthesis because you can see here opening it on record as having its got a spawning closing parenthesis. But you can see the ear, we're having this closing curly bracket, that is ischemia not having its goddess morning opening curly braces and record this particular expression is not balanced by not having a balanced parenthesis. So this is how we will distinguish actually. So we will ask user in our program to give the input expression here and this, considering this example to explain you, the diagram shows me we'll make our core flexor relax the user. To give any input exploration. You'll see how to do how to write the instructions for all of these requirements that we want the user to give the input expression. And then we will write our program and we can check whether there's been no expedition is having the balanced parenthesis or not. What logic we should apply your so let's see here. For instance, stack is empty amino then we are having the thought, which is at a reference point because we will be performed with the help of copy, will be able to perform which involved at the top of the stack. So therefore, the top will be neutral equilibrium minus one. When the stack is empty. This is stop using EDI. So this is the input expression. So for us what we will do, we even scan out of wall expression from beginning to end. We reject. We will check one by one. So we will simply consider immunity can edit for this input expedition. Eddie off character. So in that case, we will be taking a loop and we will be moving from beginning to end. So one by one, we will first off all rehabilitate these elements source. And you can see here the 0th element of this array, which is a character. It is having this opening polymerases. So this is whenever we are getting, we will make our logic, we write a logic when and we're getting minima weekend. But opening curly braces are opening round record are opening back and record. These are all opening bracket that is also the left call, this left parenthesis. So whenever we are getting different parentheses but an opening parenthesis. What I mean by opening, opening means this known record can be opened. This curly braces is this dipole parenthesis is opening curly braces. This typo emphasis is opening to co-vary red square brackets. So whenever we get these parenthesis, then we will simply push it enough stack. We will maintain the stack that is relevant in the stack bought electing these areas of really maintain the stack in order to put all these parenthesis, which I did. That is opening parenthesis one by one. We will be scanning from beginning of this, the Leanne and whenever we aren't getting opening and we will push it in stack. Once. This is one of the logic which we will put. So why we are putting like this? Let me tell you. So see you, you can see here initially we are having opening curly braces. What we're supposed to do, you simply push it. In fact, let me modify the stack. This is what I have modified when I get this opening curly braces. So I will make a logic whenever I get any opening parenthesis, I will simply pushing the stack. So you can see initially the stack was empty, sort of was minus one. We know that. Then We are watching this. We want to push this element that is opening braces because we have to make the logic whenever we get an opening, parentheses are also called as left parenthesis, then we have to simply push it on the stack. So yes, we are at the 0th element of this area of characters will be here. The 0th element is curly braces. We are scanning from left, from beginning to end. So we add at this position. So we got Collie opening parenthesis, so we have to push it so we will simply implement the top. By van. We have seen how the element using MOS stack already we have written the program and executed in audio session. I recommend, you could recommend you to go through two sessions to understand how those stack using that evokes upward. In that case, for push operation, we want, which is going opening curly braces, we will simply increment both i months. So it was minus one becomes 0. Then this was eaten or at a given simply store this element that is opening curly braces. Now next time when we are at this next, notice we have scanned at 0th element of the array. Now next time, we are not bothering about pi because we have to just check whether because the stack we have to maintain for not opening parenthesis. So now we blink again, we will not do anything again, real disposition. Remember since we are moving from left to right, that will be under multiplication. So again, it is not, and it has nothing to do with the parenthesis. So then again, one more iteration will happen. That is, and we will be at this opening bracket. You can see here this opening round brackets. So what we're supposed to do since Mia, our logic is never got any opening parenthesis me I said we had indicated that it is, so don't record or whether this curly braces, whether it has square brackets, we're simply put, we are supposed to push it on the stack. So now how the stack will be modified? I have modified my stack now on the top will be incremented by one more position. Soon on top will be equal to one. And now we will, we will be pushing this opening round racket and this tax on. You can see we are having total two elements in the stack. This is how we loop proceeding. Next we will again scan. That is, we will move further. In the next iteration we get 20 numbers, so it is not related to parenthesis. Again. An exploration will start again. We will be moving to next element and it is plus. You can see her surplus is also nothing to do with pen. This again will move to the next weekend also when we have we're not supposed to do with cool. Again, we move, we get closing parenthesis, no logic we have to put, if we're getting any on the closing parenthesis. So you can see it also closing parenthesis. Parenthesis closing. So this will be the losing, this will be the closing parenthesis opening bracket, opening curly braces. And the opening squared, sorry, what will be the closing parenthesis? It will be closing, a round bracket, closing braces and losing squared. These are all closing parenthesis. We've been checked. We'll put one more if condition and logic. And we will see if we are getting, we are getting from this input expression, any of these closing parenthesis. All struck while we have to check in that case that whether the stack is empty. So here you can see we stack is not empty, we are having some expression then that is well and good. But if we aren't getting any closing parenthesis, which is also called as right, but, and this is, these are also policies. This is the opening parenthesis, which are also called as an emphasis. Whenever we get any closer closing parenthesis are all right. This is our role is to see whether the stack is empty. If the stack is empty, that means you can understand that. Amino let me, I'm maintaining the stack for storing the left parenthesis. But if the stack is implemented in a vet scenario, if condition that MES seeing this tech is North empty but it's considered if the stack is empty and you aren't having any, you bought any right parenthesis, it is closing parenthesis in that case, you can understand that we are having, we either have a right parenthesis autumn water than that of the left hand. And this is because the stack is empty bed and he wasn't me. I'm in cleaning the left parenthesis under non left parenthesis. There it is, opening parenthesis, but we got right parenthesis. If there is some expression. In that case, we will simply say that he has an expression is not having the Melians parenthesis because right, parenthesis or more than that. So in this scenario actually because suppose via having the expression state like this is six plus. And this one. You can see here that we're not having me. This is the input expression. Then when we scan from beginning, didn't we get disclosing round brackets closing parenthesis and then B at what we have to check, you have to put condition, but if the stack is empty, and then you can see here, in that case, we will put the message. This is also called list, right? Parenthesis are closing parentheses. It is more than debt of millet. You can see there is no lipid and disease, so therefore, this particular expression is not having a balance parenthesis. This is the scenario. We have to also check that. But in that scenario, what is this situation be gone, this opening, round brackets that is also called instead right? At anthesis opening and closing condition we are getting this closing round bracket, which is also called as bread increases. In that case, we have to pop. The stack is empty or not. Stack is not empty of busy because we have put all his opening parenthesis on depth will be, will pop up from the stack. You know that dependent I'm gonna be called upon to not pass and we cannot pass any element. By default, the last element which was, which will be formed very top is pointing. So top is pointing to the last element. So therefore, opening parenthesis will be. What is this situation in that case? You can see here that Let's pick this node. But you can see here, we are at this point. So this is that we got this. Whenever you're scanning, we got this parenthesis. This is called, let's say closing parenthesis, middle, name it like this, and output. We can also name a button that says any name we can give that. We should able to understand what is the purpose of this spread emphasis, this parenthesis, what we are having here, when we're scanning, we are getting this round brackets. So this is the round bracket which we are getting. We will write a complete program. Don't worry about that, just to explain you writing. So whenever we get this, we had the first thing is that we have to check whether the stack is empty or not. You can see now the stack is not empty, so we as opposed to four. So when we pop them, what will be bobbing? That case, this opening round bracket will be formed and we will collect it and you'll collect it in the variable. Let me, here we will collect in the open. This is meaning that we live in like this, or we can name this is m, this naming on with Alonzo clear parenthesis. What other names for that? We get when we bought what we get, we get one. We've popped from the stack. We will get this opening round bracket. Let me modify the stack. So this will be the stack that is the opening them. Round record is Bob the ectomy and we have elected it and we say opening, but this is what we are getting this. So you can see dog-gone decremented by one and this referring to this previous element which is there in the stack that is opening and residual. Now, let me recollect we are having this particular closing round bracket and we are having this opening round bracket. Now we will, what we will do, we will put the logic. You write some function, one function that didn't mean that these two inputs, one in progress, this closing parenthesis. And I let him put best opening parenthesis and we will write, will pass these two inputs to one function, which we will call it as much but indices and which Mill, what we, what is the focus on gratitude? Check whether these two inputs, whether it will check whether this particular is having the corresponding, it is, it will check whether this is having its got a spondee opening parenthesis. We will have different scenario because we will have for this round bracket. Check on that. Even we will check for the squared record after the opening squared record is having its got his funding. Closing square bracket rather than opening curly bracket is having its got his funny, closing curly braces are on the end. If it is having, then it will return true. It will return one. If it is not having, then we'll return 0. That is false. So that means it is mismatch and we will give the message gender expression is not having balanced because this is what we have to check that we are having whatever we are hoping we are getting. It's a spawning. We're storing. First of all, the scanning is this one. So it should have, it's got a spondee and emphasis. So there it is though, a software program right? Now, this is how we proceed. So you can understand many bonds, these two inputs and we put in the function and Matt will see what is the logic in our program. Once we rank the programming code block ID, then it will return one because it says it's got a spawning this, you can see here in this opening parenthesis as Hamming code is forming closing, so this will not, and therefore it will return one. Therefore, we, we got the message, yes, it is having violence. Now we're still not done completing the whole expression. We are left with an adult. I didn't meant to be scanned. In the next case, what will be scanned actually now what we are appending with me or friending with, and he was healed and we appending with this last element of the array and that is the closing curly braces. So when, so again, just moving to the Viagra. Yeah, definitely. You can see here this closing curly braces we are left with. So we add at this point in the next iteration. So now again be gone. Losing. Recognize what is what logic me and upright minute when we get to closing curly bracket or blue and the closing parenthesis, which is also called as right parenthesis, then we have to simply see Foster called the stack is empty or not. So in our case the stack is not empty. Then if not empty, then we have performed. Then we pop the value which we get factoring. We will be collecting it in this medium like this opening fed into two. So let me modify this FAQ posts. This is my modified stack. That is we have for this opening curly braces and now disability modified to this. We have an endoskeleton was this one. We add a disposition that is mirror the last element of this expression. We are at this aluminum. So therefore we know that this one is not, I'm losing the reason this is not going to be gone, this closing parenthesis, then we fall from the stack. What we collected it in this variable. And so this is the bulb demand that is opening, opening curly braces, and this is the closing curly braces which we reviewed. Beyond scanning, we added this element, yes. So now we will put, we'll call the function mat and pass these element and this element. And we will check whether these are having it's good as bonding. But in this method, this opening parenthesis as having a spondee closing parenthesis in our condition, yes, it doesn't have any depth would be written one, the match function and return one. And level will display messages. It is not balanced parenthesis and then we would see spill any element we got left with food scandal. Now we have scanned the complete expressions. And what last task B as opposed to do, we are supposed to take. Our stack is empty or not because here you can see since this is our balance, so it was staggering on VMT. But walk about the scenario. In that case when the stack is empty, when we are done with scanning the whole expression than the stack is empty, we check that yes, if is empty then it will, we will finally displayed. We will return from that function. We would simply say, Yes, this is expression completely is having balance parenthesis. But this does stack is not empty, so handbill the stack will not be empty. Index position. So it wouldn't be having this expression C. Having this explanation save. Then this is an expression. In this case, you can see here when we scan this list as input an expression, this, I'm just giving you an example when the statement on BMP. So when we scan from the beginning, we've got to get this. We get this. First of all, is the opening iran brackets. So we watched it on the stack next time when we scan other element, then we get this second opening round bracket or unique portrait on this thing. And we get n, not bother then plus then 20. Then finally we get this closing parenthesis we're supposed to call mistake. Mistake is empty, no stack is not empty, so that is available. Then we will, once we bought what we will get this opening bracket. So we will call the match function yes, it is having balance. Finally again, we will see that we reach to the end of the expression nothing better, but still in the stack. Opening parenthesis we are left with. So then we can understand, yes, Nortel, balanced parenthesis because we left parenthesis or more than that. This is how we will verify the input expression that it is having a balance but emphasis on North, I hope so you bought this complete naughty before writing the pro-Nazi. Let's come to the code block ID and let's write. Let's see the logic. I will explain you line-by-line this we are writing in C plus plus and it will be executing. So this is my main function. Many new law supports to create a project in the code block ID, you need to install the board blog ID display available, very easy to install and just write the program. And then you create a project for C plus plus. And you include this header files. I mean function by iostream because I have an input and output. Then C string because I haven't used some function which is related to string. Let's see, man, we will use that and this namespace that isn't using namespace, std. Now I'm defining an extra 20. You know that in case of stack using edit, that doesn't mean I'm creating this tech static Navy have full size is fixed. We cannot change the signs, so we have to give this size on this statement we created at the initial of the program. The program is executed. We cannot change the size because I use is fixed. But there is a reason I'm giving the seismic and anything hash define an extra 20. And you can see I'm communicating. This is the stack which I'm creating using EDI. So how to create? Since these are all the stack, we will maintain the opening parenthesis and give n that will have on the parenthesis that is opening. So creating the era of the record. And I'm giving the size that is here is defined as 20. You can give any sites depending upon the otic Wyman and the necessary things for the stack Mino talk biotic wired in Italy, the stack is empty, so we'll initialize to optimally minus one. Then let's go to the main function and let's start from there. And the main function is supposed to follow the entry point to start an execution of your program. So whenever you execute your program and roll thumbs, this main function, nine by nine, these instructions will be executed sequentially. What I'm doing in my main function, I'm declaring one Eddie, that is all the data type character. All elements will be of data type character. I'm picking this, this is what input expression. So in this Eddy, I will take the input expression and ask first of all, the user to give the input expression and collected in this edit. Let's see how to use it. Video we'll be, we'll see how to make he was offered balanced. So this is just a display message on the console. We UCL in case of t plus plus as Howard the default, if you are writing those same program in C code. And then input and output support imprinting on the console. You don't use SEO, you use the print function. That's only the difference of wool, then you can simply copy paste if you are writing and C language, just make sure Amanda, you on using CL, your sprint depth and whenever you use, CAN use scanf and once with the header files which are there and you create a project policy approved, see project Nazi placements and you have to include instead of these header finally, you have to include stdio.h and Europe to include the necessary how to find a string voltage. This namespace std is not required if you write the programming language. So this is just am just mentioning you if you want writing before moving to the same values. Yes. This is a display program to check whether balanced parenthesis. Now, one more display that is the entire expression and guinea parenthesis all what I am getting this message so that he was able to understand yes, uses opposed to enter the expedition. Therefore, the CEO and then user will get an expression which I have called this function that is, so you can see I'm making the use of string. It is, is the function button. We will ask user to give the input and get their input on the unsold that will be a string. So therefore we are using get SS for string and this EIN an XPS. So I told you this year we'll collect the expression, input expression. So we collected in the string by the name, which is a character array. So this is an input expert in intellectual exploration that is, in our case, this ion expression. The user give this whole expression, we will collect them that Eddie off, Correct? No. Then I'm calling this function check bed and this is all what this tech bed emphasis do. Foster quantity in this check parenthesis being positive input expression. You can see and you can understand what we will do. We have already discussed mucin that given a different, different logic, if we get any left parenthesis, then we are simply SF wants to push it on the stack device it began to write parenthesis we as opposed to check if the stack is empty, then we will say right parenthesis are more than the left parenthesis status stack is not empty. Then we will see when popped from the stack and we will simply pass it in one mode, functions a match, and we will see it as having its got a spawning parenthesis. So let's first move on to this function and see why. Return visitor Tony. Yeah, collecting the return value in this video, but in orders is balanced. This is not an integer types of discipline. This return me Yvonne. A dysfunction in return. I know uneven and that it is a balanced. So this is the reason I'm collecting here so that later on I'll make use software will just display whether the complete explanation is bent into the narco. Move on to detect. We'll move what we will see. What is their definition of this check at anthesis when we're passing this whole input expression. You can see here this is the complete function. Object parenthesis here. This is the whole function. So let's see what we are doing here. So here we'll have, we'll have the return type one since I've worked with Ribbon 10. And the input will be in preventing human sad Adi, which we are passing the input expert. In fact, we are putting, notice the whole expression that is this exponential we are passing. Now. What we are doing, we are taking a for-loop. Why I'm taking this Guadalupe? Because you know that we are supposed to scan from this beginning of this exploration, that is from this. And that is a reason maybe required upon loop. For loop int I is equal to 0 I less than spring length of this input expression. So how much time the iteration should happen and at the end of the string so that hold this I will start from 0 until n. We want to rotate and then I plus, plus this inner for loop. Then we had, I think that logic that is one-by-one each element we've already tell him and he will change these conditions. Notice even check this expression subscript I is equal, equal to this logic is what it is done logic or to check it, we are getting element as our opening parenthesis on. So quality is a left parenthesis. So you can see I've put this event, you do EXP and this is the subscript I. So here I initially will be 0. So therefore, what will be here? It will be this opening curly braces. So we have a deck. What each element we'd have to check all these things if we get any opening curly of racism, if we get this opening round bracket of Rigoletto opening grand record and what we as opposed to do, we're seeing me as opposed to simply watch it in this stack which which I have for you, that is the purpose of the staff to store opening parenthesis or left parenthesis. So we are simply calling which function on the stack since Vietnam blimp using the stack here. And we're passing this opening that is EXP S subscript guys want subsidize, want to detail. In every XPS, subsequent 0 will give you the opening parenthesis I'm passing in. I'll call the push function and simply pass this opening curly brackets. So what you want, it'll ask is to push this guy opening curly braces and the stack. So let's see the logic for that. We know that it's a simple audit. Me have already seen in our earlier session for the stack which involves using an EDI. So this is assistance or why I am taking input or human because I'm passing opening curly bracket because our character, so I'll select in disconnect this and that. First of all, before watching any element in manual using an EDI, we have to always check my than your stack is full or not. Second spoiling, you are not supposed to push my stack or whatever and ago, so this is a definition. What is the definition of stack will be known. This is my Instead. If this is if top is max minus one, we have seen that whenever there is a reference to the element which is the last element, which is simply the index is like this. We have defined already the size of the stack. And if we add at that size, that is the last maximum size, that means the stack is nothing we can poach already been written by an impact is indicated. If the stack is not political returns 0. This is, we have to check men, it will be called me an MMU, Bush before pushing any element. In our case, the segment on people. Because the earlier at the starting, starting it is now then what I'm doing, I will check on this tax. So what does the logic work watching we have to force implemented top. So here I have written this the same logic in one line. It says, so i'm I'm using the agreement to lipo postdoc will be incremented and then the stack underscore left parenthesis subscript will be implemented from minus one to 0 because it will be incremented by one. In Italy, stack is empty, so women may agreement thought becomes 0 minus one mutually definitely get stuck under school penances. This is subscript 0 is equal to this value or table users by alphas in equilibrium. Finally, in our stack we go on to opening curly bracket pushed. This is dumb thing which we are doing here, then moving and the next, this check bed enthusiasm. Then this is awesome. I'm just matching the dog scenario and logic to check whether an element is having is Zope link button business logic one. So we have to put because different logic and we can also get the closing parenthesis. So this node, this if condition I put another loop itself. You can see that if this input Alignment is the closing curly braces auditors. You can see I'm using the auto period of you because it can be anything audited every result opening or if it is a closing bracket or if it is often losing squared record. And you can see here how we are checking closing curly braces or losing ground bracket of using square record in that case, what logic we have to do, we have seen that whenever we get the closing parenthesis or we can say right parenthesis, we are simply supposed to check, post very posting the stack is empty or not. If the stack is empty, you can see in that gives me will flip a message, right parenthesis more than their parenthesis, this message given game. And then we will return 0 on. But if this stack is not empty valence, good, that gives me will simply pop from the stack and we will collect in the variability that is left parenthesis on which is also called list, which is returned. You can also name it as opening parenthesis. So you can see why via hoping. So first of all, before we get started in this video, but is that correct or no? As I told you, we will call my function and we'll ask this left parenthesis that is opening curly bracket here. And what does this expression subscript i, it is Mozi, it is simply as S2. See how so then what we will do here, left parenthesis needed Bob. So whenever we are having me economy in the case, actually when you want to get the closing parenthesis. So first off one, in our scenario you can see that in each element that is opening curly bracket, we didn't got any closing parenthesis. So this condition, first of all, in our case will not be satisfied. We will read this condition will only be satisfied because we got this opening curly brace, curly braces, and we have pushed it on the stack. But this condition will not be satisfied since it didn't. We already heard this conditions are despite. So therefore, this is just to explain. You will not even put different, different conditions for each element. Then it is closing parenthesis. Then we have to simply check whether the stack is empty. The stack is not empty. Well and good. Then we will pop from the stack will be stored in this variable, and we will pass this left parenthesis and this left parenthesis will be cut as will be the parenthesis which we are pumping from stack. And this scan parenthesis, which will be there. What we get as a closing parenthesis. That would of the things we will be asking to this mass function and what we will be doing in the match function, as I thought UV will be simply. This is a match function which is done in vitro. You can seek the area having to input that humans left and right. So here we are checking if the left parenthesis is, This'll opening curly bracket braces and we are using angle for either because then we know manuals, me it was an operator lender to off the condition should be true, then only it will return true either or it will return false. If my knob, the condition is false, it will return false both and each endpoint it will return false. It will return true. This one condition, and if this other condition, both of them are true. These things, we are checking whether this parenthesis as having its beta's forming identities, then it will return one else-if, if it is having this opening is having its got his funding close, if this opening squared is having corresponding R-squared record, then return one. Otherwise, if no, none of these, then it will return 0. In our case, this condition is not satisfied because we are opening at the very beginning of the expression and we got the opening curly braces which we have pushed on the stack. Then this is what the logic we have two active I didn't want will be the scenario again. Scanned and excelling meant that despite so we have to not bother. The next will be multiplication, not while the, again, we had this opening it on brackets. So we will simply, this condition will be satisfied. That is, you can see this first condition, that is this one, because we are using a court operate in order to operate them enemy use any of them is true. The world decided will be it through. So therefore this is true in that case, so we will simply push it into stack. So in the stack you can see me half cost curly braces opening stored. And the next we have stored this opening bracket. So we are having two elements to totally top is pointing to this last one. Now, this condition will not be satisfied because we go into opening bracket. So therefore, again, next iteration, what will happen to get 20? Then we get less than, we get four. We are not bothering about that. Again, we are getting, we are getting though laws, bed and thesis that is also for this, right? Then finally, what we are, what logic this condition will not satisfy. This condition will be satisfied. This condition will be satisfied. Therefore, because this particular thing is turning true. Therefore, condition will be true. In that case. That, that is, since we are here, therefore we live simply pop from the stack. So when we pop from the stack, so fast upon me I protect causes the stack is empty or not. So in our case, yes, the stack is a stack is not empty because we haven't put any men's prison. And the last element we are having this opening round brackets so that whole stack is not empty. This foundation is not doing our case. In Xcode, we will be simply poking from the stack and we will be collecting in the variable thickness left parenthesis. What we have heard in communicating the opening round brackets or openly and following the match function we're passing this opening round, recognize his name is emphasis as well as me or passing this scan. So what we are scanning, scanning this clo 5. Infix to Postfix Conversion & Evaluation of Postfix C & C++ 1stHalf Session4: Hi, new session. I think the program and executing it in Windows infix to postfix conversion. Evaluation of the postfix expression which we obtain after converting infix to postfix the same. We will be evaluating it in C and C plus plus. In earlier sessions, a veteran in the earlier two sessions, I have covered what infix to postfix expression. We have seen that. And we have also seen how to evaluate that is postfix expression, which we have explained that the same on the whiteboard, I will be taking the same piece, same program which I'll explain in the whiteboard. I will be, I think I will be executing it in the code block and we will be seeing its output. So this wouldn't be a good practice for you, so that you can simply copy, paste the whole program in the code block ID and windows and executed. So we'll be seeing it in C and C plus plus. In this manner, you will be having the hands-on to already the concepts may have seen in the previous session on the whiteboard. Let's write the same code in a code block and executed. We will be taking this expression that is the index exploration. You have already seen what is in fixed expression. You can see here the infix expression just to quickly revise in which the operators you can see here plus minus all these other operators. This is present in between operands. In fixed, that means the op, depending upon the position on top later if they decide what will be the expression. So why this is called as in fixed expenses. Imagine the reason this operator is present in between the operands. So you can see here in-between Ethan for this class is present, in-between important with this minus is present. So that is the reason it's name is in fixed explanation, not heal. The same explanation. You'll be writing the code. We have already seen the code. I have explained you on the whiteboard in the earlier session. In pickup, the same code on the code block will be executing it in C and C plus plus when NBC in both languages. So let's move on. Let's write a program and executed in those using C and C plus plus. I'm just moving to the moving to the code block. Phosphate will be seeing the code C plus plus. This. Let me increase the font size. This is technically I have created in the code block ID, so we know that it is very easy to download and install the code block IDE. So you can just refer the same. And you will have the ID wherein you can write the program and executed. So this is the C plus plus program for conversion of infix to postfix and evaluation of it. I'll explain your line by line and you'll be executing it. I already, I have explained you on the whiteboard, so I'll just quickly revise. I just wanted to show you how does it execute. The same program which I have shown you on the whiteboard, has been an explanation how we will do the same program exits be executed on the code block. So we'll be also seeing it. I would put on the code blocks or this is this will be in helping you to have the hands-on on the same program and you can copy paste the same code I have on school. And there'll be also I have also shared the same code. You can simply copy, paste the same code, and execute it and see what will be the output. The first list, let me in the notepad just to quickly explain you what in fixed explanation we will be taking here. Consider this in fixed expression. You can see here this end fixed explanation we will be in. You can take any explanation. Yeah, I'm taking this in fixed expression, this seven week on voting in postfix. So first of all, you know that how does this in fixed expression should be converted to fix. So as we know that the parenthesis, first of all, fixed explanation, who pulls fixed, how it will be fixed will be converted to the postfix. So it will be scanned. It right like this. Starting from this. It will be, it will be scanning the expression from left to right Foster point. What will be the thing in this case? What is the purpose actually, using the stack in converting the infix to postfix. We have already seen it that we are using the stacks, but FECA and cheap purpose because it will not take time. You don't need to scan again and again. Expression we have seen if you don't use stack, then you have to scan the whole expression from left to right like this. You have to scan from this, from this. If you don't use stack, then you will be scanning from left to right and you will be seeing, yes, this parenthesis is present. So you know that parenthesis is having higher priority than you will. And it should be evaluated first and afterwards. Again, after the evaluation on that again, they will be spanned from left to right. Again, it will be checking for next highest priority. So in this manner, if you don't use stack in this minute will be executed. So the reason why we are converting this infix to postfix, and we're using the stack because for a DC10 purpose, so by using this tag you will have will be converting to the postfix expression. We have already seen in the previous session by what is the fixed exploration? In the exploration in which the operator is present after the operands. For example, if you consider this exploration, that is eight plus four, this is an expression, so it's for fixed will be 84. And then, you know that why we are converting this because in this manner in postfix explanation it also remove the parenthesis. The parenthesis are removed, as well as the priorities are arranged in a sequence order. So that is a reason you don't need to scan from left to right again and again simply by von go from B to convert infix to postfix. That result in postfix expression will not have an instances. In digital operators will be arranged in sequence order. So that is a reason actually we are converting to postfix and that can be possible efficiently using the stack. So here we'll use the stack. So I've already explained all these concepts in detail in previous session. I explained you on the whiteboard. So here this means, this main purpose of this session is to just show you same piece of code I've written in the Windows. And we'll be executing what I have written in C and C plus plus, you can simply copy, paste the same piece of code and executed. So just to quickly revise the concept, I mentioning it again, again for you friends. I hope you understood what is the yield. Finally, we know what we're supposed to do, what it will do. We have seen that the stuff on this parenthesis, we know that what will be the logic? First of all, we will be scanning from a to B is the process by NB scan from left to right. First of all, we are using the stack. So what we will do, we will, we will simply store the operators whenever gonna be scanned from left to right. In this manner, we've distorted the operator and we'll take one num eddie. That name will be the boats fixed at it. We have seen that. And we, we will be taking a stack. So we have seen waterfall input required in order to convert this infix to postfix one is required. Which name will be the pole speak, Sadie and one stack using the EDI, we are using the index. We will be storing operators. We have seen things are required and previous session. So first of all, we'll scan from left to right, and then we convert infix to postfix. As we find operator, we will store in certain array that is a false fixed. Then we scan the next symbol. That symbol is plus. When we find the operator, we will simply push it onto the stack. We have seen that again, the operator comes for Rivest will simply store it in the area. Again, the next operator comes to what is it? We have seen what rules we should follow when we convert infix to postfix. First of all, these adults symbols, these each symbols which comes. First of all, please keep into mind that whenever you operate your CMS, you as opposed to store it in the postfix setting when it comes to you as opposed to store it in the postfix setting. But when the operator's comes, you have to deal it with best stack using the array. So first of all, you have to spend never looked at this is the same scan symbol and you have to already, we have seen all the conversion rules. So once you get the symbol is suppose plus. Your stack is initially it is empty, so you will simply push it onto the stack. And other symbol comes. This is an operand. This you have simply storing, added again another operator. And say, this is a scan symbolic, that is this minus if it is priority, if it is lesser, as compared to that of the top of the stack. In the stack you're storing if there are some operators. And you know that what is the top of the stack validity reason. Decent element is your top of the stack. Obviously in this tag we're storing the operators, so decoded top of the stack will be operator in the priority of this fence in ball, this is a scan. These will be the scan symbols and you compare it with the top of the stack. If it's priorities less than the top of the stack. And you will simply bark the top of the stack and you will store it in. Eddie. You are suppose to compare against the same scans involved with if you thought I was scared Bob, the top of the stack element of this type you will form, which is the topmost element of this technical and meet up. And it almost element will be formed. Then the next step it will be there thought will be the other previous element. So again, you have to compare these scans involve embedded some operators left in the stack. Then you have to compare with the other priority of that symbol with that priority. Again, if it is a priority on this Canson ball is less than alpha of the stack. Then again, you have to follow up on this accident. They simply keep on pumping. You find this priority of the scans symbol lesser than that of the top of the stack months you'll find this top of this, this scans in bulk rarity greater than the top of the stack. If it is like this, then you simply pushed onto the stack. So we have already seen that in our earlier session. If you find the priority of the top of the stack. If you find the priority of the scans symbol equal to the top of the stack and you have to check the associativity. Associativity on these operators are from left to right. That is, you have to have football up on his thing. But if it is right to left and you have scanned symbol onto the stack. So these are the rules which we have already seen. Also I will be making, I will be also mentioning these rules in some documents so that you can also it for your purpose. I hope soon you got the point. So first of all, if this symbol, these are the scan, this one is four minus, these are the scan expression. What I'm symbols will be the scan symbol comes with operators. Then we have protect added emit back-up bus stack, top of the stack. Quickly devising input. Symbol commodity is lesser than you have to pump the top of the stack. This can symbol priorities greater than. You have to push down symbol onto the top of the stack. If it is equal, then you have to check the associativity. I did if it is left to right associativity. Again, then in that case you'll be popping the top of this tag. If it is right to left associated within any good fortune. This is what the rules you have to follow when you what infix to postfix expression. So I hope you understood. This is just a quick revision we have already seen in the previous two sessions on the whiteboard. So the main purpose of this session is to just execute the same program AT, for you under the code block IDE. So first of all, you can see here this is the C plus plus program and the same code. You will be finding it. I haven't given you. You can access it and you can copy paste the same for your code. I've made you made it available for you. First of all, in C plus plus code, as we've seen the C plus plus, then we will see in the C B also executing it. So you have to foster upon mentioned the header files which are required. You can see here iostream and Nortel string dot. All these header files are required because you will iostream is for inputting and outputting when you call the functions for the same. And these, I'm defining these macros. So we will see when I'm making use soap and stockholder view here we are using stack using EDI. So notice the reason this is used to define the size of the stack so that this is fatigue. Because i'm, I'm using, I'm using a static array. We are using stack using arrays is static. You cannot creating an empty. So therefore, I haven't mentioned the size we have prevention does ice at the compile time itself. This is a size which I am giving full stack using Edit. You'll see how Nika you saw fit. In case of this C Plus Plus, if you aren't aware of It's language you will write. You can write the same code in C plus plus. You can understand it. If you want to read off see, I will show you how to see also how to write for C plus plus. You know, we are creating, we have used up to us. And there are different specifiers public, but I read all this. This is my class named as infix to postfix. You can see here this classes like this. This is my class infix to postfix. And you can see here this is oblique axis by a specifier. And you're a private access specifier. So far, so far, in public access specifier, you can see here, I'm creating the constructor which is public in fixed reports, fixed construct, you know, in constructors initialize the video amendment of getting uses. You can see a public I had made all the functions which I'll be using in my program. I had made it as public and I had made private for an all member variables. You can see here we are using stack using an array. So we are using the top will be referring to the topmost element on stack using EDI. And this is not Stack named by elongating stack max. So this is the stack that didn't. We will be storing OR operators. This, I'll let you know what is the purpose of this. Extra things that is whitespace. So what I use here in our class. We know that since we are using stack, using Editor VR, we're supposed to know the operations related to stack out of Fortune fall. Why does push and pop is used? We know that we will be popping at some condition, may be pushing at some conditions. What are the rules we have already seen it. I will, I will show you how to prevent even use this. And this function that is in this function is flawed and working infix to postfix explanation, since this program is for conversion and to infix, to postfix as well as the evaluation of the postfix I have on social and new on the whiteboard, how to evaluate the postfix expression. Therefore, this function will be just converting in fixed to false fixed, which I had to buy the rules which I mentioned. And the another function, evaluation once we get the full speaks explanation, dysfunction is used to evaluate the same postfix expression. And then priority. This end disparity incoming by these two functions are used because as because one function will be called when dealing with the left-to-right associativity. We know that bottom left to right, left to right when associated with the comes into picture. And then vitae, your priority of the scans in bold is cmd and same to that, all the priority of the top of the stack. Then associativity comes in, index is in the associativity is left or right. Then you will be popping the topmost element of the stack and you will be supporting in your result in postfix expression. Associativity is right to left. In that case, you will be pushing the scans in bone onto the top of this stack. This is the thing, you know what the in-depth case, the left or right associativity changes, then the rules changes. So therefore I have created two functions. So one function name already another function for anonymous but editing undisclosed incoming symbol depending upon associativity, whether it is left to right, I will let you know how it is executed, is empty to check whether your stack is empty or not. Since if your stack is empty by this condition is required because you aren't popping right? Whenever you fork, you have to also make sure that some elements are present on your staggered. Nothing is present on your stack and sending us for being done. In that case, the problem can come, come. So that is the reason you have to check. But if your stack is empty, there is no question of hoping. So that is the purpose of it. Now, menu, so this is the, these are the functions which we have seen water in our class. What member functions are used, what member variables or use. Now let us go to the main function and then let us see how the execution takes place. We know that in case of the main function, the control comes in the main function and line-by-line instructions will be executed. So far as tough on this control comes when you execute the C plus plus code until all comes here in this main function and line-by-line instructions. So here what I have done, you can see I had taken an area of character Eddie for an infix expression and corrected edit for postfix expression. If I am taking the character a report in fixed, we know that in fixed explanation will be simply this one which I've shown you. This whole exploration. I'm taking this up. This is my fixed explanation and the resulting, when I do the resulting expression which I should get is a postfix explanation. So that should be also an ID. That is the reason I've taken them at ease minus the input expression and the result in which we'll be getting after I Security in this code will be the postfix so that it doesn't result in the area. Now this is just a cout. We know that in C plus plus C out is to display on the MSc infix to postfix and evaluation on the postfix. This message is just to display on the console. This message will be displayed so we use c out. There is a reason we have included the header file hash include iostream. Then I'm using long int Val. The top I'm using, I'm initializing my top four minus one. We know that initially the office and the stack is empty, so therefore it will be minus one. The opposite. Bill in the mandible, you will start filling your stack using edit and your talk will be incrementing. We will be seeing that so far. Stopwatch alarm, this message will be there. So this is all explanation I have already explained you on the whiteboard. This is just, I quickly move actually and executed fast. So first of all, this message enter that in fixed explanation so that when I'm using this function getters, getters, this new school in the input from the user on the console so that you can give the input in fixed expression like this. You can give this expression here in your console and you execute the code. And then what I'm doing is I'm creating an object of this class. So this in fixed reports fixes the same class which I have shown you now. You can see here this class is by the name and fixed to force fixed overhead. And I have already given the construct the member variable, right? You can see here I'm creating the object of the same class by the name into forced by this. This is a static object creation. That code, I'm not, I'm not doing the dynamic relation. You can see here this is a static object. So by this object, I'm calling the function that is infix to postfix. And I'm passing this infix and postfix in fixes more infix expression. First of all, it is the expression which I am giving you, the one, whatever you give them fixed expedition, you are sending that and you are giving sending the postfix expression in Italy discourse fixed explanation is having nothing by this function. Whenever dysfunction via we'll be putting the logic in it with a postfix expression. So VS simply passing nothing we are passing in its postfix in fixes them. Input exploration which you are taking from the user. And this force will be nothing initially. Dysfunction. We'll do the load logic, we will do the logic and dysfunction in order to convert this in fixed reports and we'll fill in this function. It says, let's move on to this function and see what will be the definition of it. This function that is in fixed supports fixed function. You can see here in this function in fixed, false fixed, you are passing this month string and then you are passing another string here. In this exploration, in this string poster hall in this function, whose strings are there, this will be your input in fixed explanation and this will be initially no stealing anything. But in that logic we will keep on storing and not postfix expression. That by Ethan wanting to the infix to postfix. Let's see how far I stop on in this function. You can see I've declared initially the variables we receive is the use of each video, but this one by 1. First of all, you can see I'm using a for-loop here. I've already explained you on the whiteboard via, via using a for-loop because we will be scanning from left to right. First of all, this is fixed. Exploration is stored in the activity, so v will be one by one to each symbol is weekend. Compare actually with the rules, if it is an operand, we will be simply storing in the postfix expression. If it is an operator, we will be dealing with in this stack that we have already seen. So therefore, we are using a for loop. So all these non-peak I've already explained you in the earlier session. This is just to show you how we do it and we will see the output. I'll just quickly revise. So therefore, this for-loop we are using for the same string length in fixed explanation, one by one if VS scanning and this infix I, collecting and symbol. And therefore, we are using this switch case here because we will probably will compare with each routes. So first of all, why I am using this condition for using this whole switch cases, for such cases written in if condition. What does this condition? It is not about whitespace symbol. So therefore, you know that this is the symbol plus is the same on all these other symbols. We are checking. This should not be the whitespace. So since we have our main concern is to deal with the operands and to deal with the operators. And if, suppose that he was going like this, like this. So first aid comes then you can see there is a space, right? There is a space. So we have to ignore the space. We don't need to. We are not having any loop all of this space, so we have to ignore this and then we have priest scandal, a bit operator. So for that purpose, this whitespace function is used. So if you see here in this whitespace, what I'm doing here, let me, this is simple, I'm checking for this. You can see I'm created this Brightspace in which if it is a space that will be like this, I'm checking if it is a blind audit because I had if it is a blank one time, then it will return one. That means there is a whitespace, so we have to ignore it because we don't need to scan. They didn't know what that network I'm writing the switch case in that whitespace. I'm taking. If it is not whitespace, then only we are uncertain protect with the rules. Otherwise, we have been not going to ignore it because our space, we don't ignore it, proceed with other symbols. So that is a reason this condition instead, I won't tell you under responsive know this whole concept, not switch cases that are in it. And so, you know, whenever we get company parenthesis, we as opposed to a pool check on the best path. Then wherever it is a closing parenthesis via fault, all the elements from the stack, they'll find actually the opening parenthesis on this logic. We have seen the same thing, the coding I have written, so I've written the book and you can see on my one thing that is, if we find this, then one-by-one bumping must be at checking if it is opening and we have to stop pumping. If it is not opening, then we have **** off. And this we will be storing the result in postfix expression. This logic we have already seen Na plus minus multiplication, division mod operator so far. All of these, we're not giving break, that means this, you can see this logic will be there for all these operators wherein there is no break. You can see this logic is for the same whenever you come with the operators. Now you can see I'm taking with what I'm doing here. I'm checking. First of all, first of all, I'm seeing that men have and I'm getting any scans symbols. So if my scam symbol is this one, gap, if it is plus, if it is division like this, if it is like all these operators that these are the scans and bolts. So first of all, I have to check these. Can I get any scans symbolic projected with buffer edit you off the top of this text. Before checking the priority of the top of the stack, we're pulling this check whether they are stack is empty or not. There is no operator president, there is no question to check. This condition is must you have to check whether the stack is empty or not? If it is not empty, valence would then only you can check the polarity of this text. So you can see here I have checked, but I didn't feel good stack view on this Canson ball, if it is lesser or equal to that of the top of the stack. In that case, you can see here, I'm simply. 6. Infix to Postfix Conversion & Evaluation of Postfix C & C++ Session5: Fraud before these operators that it's four plus minus multiplication, division mode ratio. You can see I have not written directly. This instructions may have written. So on this instruction which is there, will be executed, will break, will be executed for the same event. And bird, it's embolus plus or minus on it. Itching, division mod reach to support all of these cases wherein we have not given the same instructions will be executed. The instruction that is this, what we're supposed to do? So whenever the scans in bold is plus minus multiplication division modern raised to the same instructions will be executed since we have not given brake. And what is that instruction? First of all, we know that whenever the operator comes, we have to check did not rule. That is we have to change the priority of the scans involved with the priority of the top of the stack. That hole. Once this operator comes, that is, is our exploration in flux expression. So whenever we get the operator notice plants or it begins minus bigotry is true, then these other post-op follow me, know that the school in Victor expedition in which, in which a business in bulk glasses are symbolic for deserts and minuses inborn and so on until the end. So if it is an operand, we know what we're supposed to do, reassembly supposed to sort it and was fixed at a, which means we have created. But when we get the operators like plus, minus, minus Audre student, we have to deal with the stack, we have to operate with the stack and we have to check the priority of this scan symbols that is this with that of the top of them. So this is what we have done. Now as you as I told you, that the priority of the scans and voltage, if it is plus this is a scan symbol. At this point. If he added which point under scan symbol, you can see the split. If the scans and ballplayer D is lesser than that of the top of this stack and the top of the stack, whatever operators present. If there is a reality of this camps and molecules lesser than that of a bit of a mistake. In that case, we are supposed to fall off the top of the stack and we have to support distorted and holds fit expedition that sports picked. If it is illustrator, if this scans and bold clarity, if it is greater than that of the priority of the top of the stack. In that case, we have simply PET scans and ball onto the top of this stack. I hope so you got, but if the priority of this scan symbolic it is less, it is equal to if the priority of the scans and monitor is equal to that of the top priority of the top of stack. In that case, you'd have to take the associativity, that is a fetus left or right associativity. And you have put simply pop the top of the stack and store it in quotes fixed expression if it is right to left associativity in that case here for some people to discuss involved in onto the top of the stack. I hope so you got, that is a reason. You can see here, I have written this thing that this condition or this condition is for associativity right to left that is radically of the scans involved with one. If it is equal to net of whatever scans and bold comes and if it is an operator, then you have to check with the priority. So if the participant equal to that of the top of the top of the stack, and we have to check for the associativity. So we know that as I have discussed on the whiteboard, that get if you're having plus Audit be having minus. In this case, you can see first of all, this flux N minus the default lesson, minus these blue dots on them or having them seem. If this is your scans and this is the top of the stack in that gives me object the priority. So if you know that in this case the priority on both of these operators or to embed CPP file mode on them. You know that if it is like this, leads to reach to this, we get this Canson bolus raise two. And if we get the top of the stack as raised two, then we know that vote on them. First of all, they are having the same priority. Then we have to deal with associativity. Associativity is what I mean if I say the associativity is right to left. So suppose you can see here, this explanation is there. You can see here. In that case, associativity in the sense suppose I'm having, suppose I'm having this expedition, see this one. It is. It, suppose I'm having, this is your expression and fixed expression. You can see here. Suppose in your stack you are having raised to and in your yard rhos cancel your scans embolus restore annual stack is also having thus top of the stack is raised to in that case, how, what the associativity, how it will be walking, it will be first of all, both of them are having same strategy because their siem operator. So that equal with an entity or equal to that associated with this dipole. Let admins execution will take from right to left like this. So first, do wrist or false, it will be two raise to three will be executed. You know that two raise to three is will be eight. Then what, whatever desert we get from the students to treat than it is in. So then what thumbtack three, then two, raise to k. This is how it will evolve. So first, what we'll do, what will be this will be evaluated at is to restrict tree. Suppose we get two raised to three. Then finally, this food is still eight will be executed. This is how it works. So definitely associated with these right and left. In case of the plus minus, the explanation is laying this minus two in this case. So in this expression so far, so follow me know in the top of the stack is top of the stack is you are saying the US and your scan demolish minus. In that case the specific gravity of this plus and minus I've seen right? Associativity, we know it. Plus and minuses this left to right, that expression first eight plus 40 will be evaluated. That is, it will be it, let's fold will be 12. Then this expression is eight plus four is 1212, minus two is minus two, so it will result to it. This is what the split takes place. So this is how it was so this is what I have written in the fall. Also know that dismissed thumbs interpreted and we have to consider the associated 3Ps rank. Or else what other operators which we have taken in our code by associativity is left to right. Therefore, I have in this condition or that condition on you have checked here. I'm checking that the stack on top of the stack, if it is having this operator, it is, that's three, it's two. And if this involved scans and roll is also these two, that is, you can see a bulldog demo damage to in that case. We know that the index is the situation. And because we have to simply what we're supposed to do since I told you that in case of this FBI having the same operators, then having no identity on top of the stack and it's going to involve same. Then we're protect with associative, associative entity associated with these right-to-left than what we are supposed to do. We are not supposed to pop your support. It'll push deeper what I have done. In that case, m equal to our left to right are associated to repeat it. This is a different meaning. We are, we are having the same priority per top of the stack and scan on the symbol. If the associativity is left to right and we have to Paul, but if necessary, abilities, right, for lambda and we have to poach. Therefore, I have created a different function that this priority incoming symbol. So in this gravity incoming symbol, what I will do Post.all sinter and supports to push me know that it is associated with these regular Piazza posts to porch. That will what I will do for pushing purpose. I had created a separate function that is disparity, this attack in this one that has polarity underscored incoming symbol. And if the parity of the incoming symbol, if it is, then that dog, dog priority of the top of the stack. So what does it mean? So Falstaff on what I'm doing, since I'm, I'm supposed to. Which this symbol which began that is, if the top of the stack you are having as raise two and this incoming symbol, that is this one. If it is all of its wonders onto, but it's still having this one. This imbalance on the stool. In this case actually what I will do, I will use variety of incoming symbol. I will return. The return value is compared to that of but I did you of this tack on the top, what does it mean to quadrapole? We know that though. Let me come here in this expression as I thought you epidemic students to do. In that case, suppose I am having stack is one. Stood as well as my scan symbol. Is also the symbol that is scandalous. This, these are the symbols, the operators which are here and then fix exponentially symbol. And we know that the top of the stack, what is the top of the stack? So it both of them are equal. What I will do, since I'm supposed to be no, the associativity for this is that we're supposed to, which the scan tin want onto the top of the stack. So that will whatever you do, I will simply have created a separate function that is the same Sima operators or that I cannot call the same function that is the priority. So you can see here In disparity function what we are doing. Let me show you the function then even understand this priority function. The function, why I'm using this, this is a class name scope resolution. And because I had created this function, remember, I have defined it outside the class. They put on using this gleam of the class. And then the scope resolution. This one, the name of the name of the function. So what I'm doing, if the priority, these other ones, these other characters. So if the priorities, that amount of time returning 0 evidence opening, if it is plus and minus, we are not using anything here. So I'm returning. Since we, I'm returning at some, some positions are not using break. Since four plus n minus, you can see partners we have not using break. The biggest note that this will be executed, the same instruction will be executed. So both plus and minus, we are returning. One. If we are having the scans and boldness plus and OH minus will return one if we are having multiplication, division ought to be higher percentage than in that case. I simply returning yes. If we stood up is we know that it is greater as compared to other operators. So we are returning. If you call, what will happen actually if you call the same function, let me tell you in that manner then you'll learn to stand it is in this case, you can see here. Yes. V-naught, if both of them are same, like the skeleton one, top of the stack without having no operator S raised to it. If I don't call out the units dysfunction if I use in this priority. And then in round brackets, symbolic rate is greater than pilot your stack on the top, I'm falling. If I call the same function, we know that in that case what will happen? It will be ignore that the parity operator is two, but we are returning is, we are returning as the editing is. For that case, what will happen to you running S3? What will happen in that case, it may be, don't you create a different country like this? So here, if we don't create rarity underscore incoming symbolically, all the same function that is priority and hence involved. And I've checked the condition greater than parity of the stack of the top, be knowing that if I call the same function, both of them will return S3. So this condition will not be satisfied because this will be returning as three. This will be also returning it's three, then this condition will be not satisfied and we will be spending more time doing. If the associativity is right to left, then I'm punchy. Whatever we want will not happen. So net board, what I have done, I have created or the print function priority incoming signal, it's in bold. Only when we can symbolise reads two. And if we get the top of the stack as the stool and only I will call this function variety of incoming symbol. And for the symbolic links to what this priority of incoming symbol, if it is greater than that of the rarity of the top of the stack, then only I will, which discounts in Poland to the top of the stack. So what is this incoming symbol? This is the incoming. You can see here. In this incoming symbol, I have written this sketch and ball and retain. It doesn't manage this and simply return for what will happen when this function is called, then this will return it or bought this symbol that is for the REI store. But for this priority function, that is this symbol when it is these two. You can see here we are returning again for this scans and ball incoming symbol. I'm returning for the Vedas or this priority for the top of the stack, I'm returning three because I'm calling the priority function that returns three. So what will happen in that case? You can see that is stuff on syllables which are row. What will happen? In that case? You can see here it is incoming symbol will return false because we have already done the cognition and this gravity is returning. This condition will be satisfied because four is greater than three. This vendor done for and disability are done three. So for descriptor than three, this condition is satisfied. Violin boot. We want the same thing. And then we simply push scans and bolded as raise two on to the top of the stack. So we know that whenever there is a race to the end, symbol and top of the stack array is two. Then we're simply supposed to push this Canson ball onto the stack. So therefore, we are doing the same thing. So we are simply good. We're creating different function which will return all the same tim volume. It will return a greater priority. So that means this condition will be satisfied and then we will be pushing. Now, once this happens, then we would simply do the break. So we don't want other things to happen. That means this will come out and then again, symbolic will be scattered. But this is not the case. That is if we are not having the scans in bolus, these two as well as we are not having the top of the stack as race to the top of the stack, as well as scansion Baldi are not least two, then in that case this condition will not satisfy. This condition will be checked. So this condition is ponder the associativity that is left to right. So in that case, what we will be taking, first of all, we will be checking that. In that case, we will be simply checking postdoc or whether the stack is empty or not. Why I'm checking the stack is empty or not, because we know that this condition, which is that, well, first of all, and this is what we are checking with the gravity. So if the polarity of the, of this gang symbol, you can see if it is necessary as compared to that of the priority on top of the stack. That is what we are supposed student, we know we are simply supports to borrow from the top of the stack and we're supposed to templet stored in the resulting postfix expression. Also, it is equilibrium. The vanity of the scans and monitor is equal to priority of the top of the stack. In that case also for the associativity, that is for bother Associativity that is for left to right, we're simply supposed to fall, we're not supposed to, which becomes water associated if the gravities are same. But if the associativity in the predators are same, specific duties right to left. In that case, we are supposed to push. But if the polarities are seeing that is here, you can see object or equal to, but here I've considered associativity to left to right and we're supposed to pump. Because what right-to-left associated in here itself, I have j and I have simply given the brakes, it will come if this condition is not satisfied, that means you can consider the associate. Fancy than that. This we will be taking also what associativity is left to right. Associativity comes into picture on leave and your priorities are seen. So if you are having the scans and boil priority lesser, you can see here we have checking lesser them that off the top of the stack. Enemy, yes, and please support to the top of the stack and stored in this result in postfix expression. But if it is equipped with, then also we're supposed to fall because we are considering here the associativity as Neptune, right? This is what, and then by averaging this while furniture, because we know that we are simply, if we are having this canceling all priority less than the top of the stack image. Suppose we have to keep on popping till we get the priority of these gangs involved. Grateful. Even if VIP disparity of these scans and ball is nestled in the top of the stack, will be involved the top of the stack. So we will then there will be a new pop off the stack. So it will be the Andrea and human which is dead in the stack. Again, we will be checking the priority with that top on this check, again, if we find the scans and more priority lesser, again, we will be posting the top of the stack. This we will be, keep on doing the same thing. In this priority will be until we get this parity lesser than this gang symbol lesser than the rarity of the top of the stack. Once this condition is not satisfied, that means a priority on the scans symbolic if it is greater than, this condition will not be executed, not both. And then finally, we will be putting this symbol onto the stack. So this is what we are doing in this case. You can see here, first of all, I had made this for access to your user. You can simply or simply checking, you can simply copy, paste the same code, and execute in your code block ID and see the result. You can simply do it for the same code, the same scene, the one cleaning you, simply medium available. What we are doing. I hope so you understood the four different strategies may never operators the scans and Walter operators what we as opposed to conversion off and fixed to fix. This is what we're supposed to do when we get this embolus. Africa's. Now the next thing is that what we have checked it now, we have to check opening. A round bracket closing. We have checked for the operators. Now the last that is the default one. What does the department we know the default minus your operands. We know whenever you get the scans involved as operands, what we as opposed to Julia simply supposed to store it in node is I didn't expect Edit expression. It is a little more splits explanation. And that is area because there is this that's involved UBI simply stored in it in this postfix area. And we will be incrementing. The people who split. This beam will be incrementing because how we will be stored once we spawned a symbol in this array, then we will implement so that the next time when we add this symbol organism, whenever doctrine comes again the same, all that's involved will be stored in debt, was eaten. And that is the reason yet we, as opposed to increment men, we store the symbol and that physician or the editor. This is what we are doing in the defaults with we know that whenever we get, let us aid or simply store it in suspects. Expedition. I hope so you want this whole thing and that is fixable, would speak to experts. And finally, since this is what we are taking a for-loop, this we are, we're supposed to do one-by-one. This for the 0th element by element taken, all this will be reached till the end that is bringing them. That is a reason we are giving this condition, string length of the infix expression of the scan on these symbols. Once we run the whole expression scan, the for-loop gets comes to the end. In that case, what will happen? In that case still we have to check that this is outside the for loop. Because one PR events or we are finished once we are done with complete scanning of the Wilkin fixed exploration ones, the for-loop comes to the n. In that case, still be able to protect your stack is empty or not. We have seen that if the stack is not empty, then we have to simply pop from the stack and we have pre-stored in this result in postfix expression. So this is the reason I've given this while loop. We have to put all the elements one by one. We don't get this stack empty and you have to simply stored in this was fixed expression that is added expedition, So that is one that will be executed. Finally, finally, you will get the result in postfix expression and we have to simply add this null character and the large don't dispose fixed, Eddie. This is how you will get your course fixed, edit, fill in this function that is in this fictional or expects. And finally, how we are calling this infix to postfix from the main function. We are passing this, we have seen we're passing this in fixed explosion is postfix was not having anything when we called from the main function. Once this function is called in Pix2Pix fixed months, this lagging strand, these are all instructions are executed. Finally, you will get the result in false fixed erythema. And finally, the span dysfunction comes to when it comes, it will go to the main function where from Merritt, what's called it was called from this main function, that is from this, from this line, it has gone from here and you will get the sports fixed edit film. And by what you are doing, I'm simply deserved. Bringing this up is to print the value. You can actually even simply display an error message if you have a blueprint for value can give in this manner. So this is what this end l is status in the new liner, this value to be printed. And then once I get this result in postfix expression painted on this program and talk on motion of infix to postfix expression. As well as the program is for also evaluation of a postfix expression we have seen in other video how to evaluate it. We know the concept of evaluation. Let me just quickly device. So from this object that is enforced into boats protect created here I have called this function evaluation of both. And I have asked this result in postfix expression, which I got from this infix to postfix. What does evil underscore post have? The input given to sports is what we got. Let us come here to the function definition of eval underscore post. It will return an integer, so this will return your expression value. What it will do. This, first of all, this infix expression is there. It will be converted to those specs exploration so that we can see what posts fixed expression we will be getting here. And that was fixed explanation. I'm boxing here in this function evaluated in this course. Finally, that postfix expression will be evaluated and you will get the result. Whatever you get the result will be deserved. All finally, the user has given. In fixed expression and in our coding vf and wanting to postfix expression. And then we are evaluating it. We've been seeing what will be the result of this. So first of all, you can see here, what is the, what is the logic for this evaluation of a postfix or whatever? That what we're supposed to do in this what is supposed to do in this evaluation of postfix. We have seen that suppose I was supposed to get this as a postfix. This is a postfix expression which you want afterwards and margin from infix to postfix. This is not postfix fixed expression, which you get. Suppose this is a postfix expression which you get. So first of all, we will be scanning it from left to right. We will be using a for-loop for the same. Once we find the operator. Once we find the opposite, it is an offering photos and plus as an operator to result. Once we find the operand, then we will simply push it onto the stack. So here this will be opposed it when we are evaluating the full space exploration. In that case, in the stat we will be simply watching her apparent. As we have seen, many moles converting from in fixed or for space exploration in the stack, we are pushing the operators, not the operators, but this is a, consequently evaluate, evaluate. When we scan from left to right, we may be getting one. And we get the sketch militia. When we get the scans and modus operandi or simply supposed to push it onto the top of the stack. Landscape. The operand APS will push it onto the, onto the stack when we find another symbol as an often again, you will be pushing it onto the stack. When we find the operator. If we find the operator, we will be popping sometimes one time when we pop, you know that recent but element which we are, which was poached very last minute, be both very fast. We know that this is a concept of the stack. So that portfolios fall will be formed. You will be collecting in some variable. And then we will be again, we'll be working full-time. So it'll be popping why we will be pumping two times so that we will collect this post for our buck in some variable and another op in collect in some variable. And this operator which we bought, we would simply therefore what we will be doing both two values. Then we will be simply operating this plus in-between this bobbed value. So that this will be, this will be the scenario. That is, suppose we get the software standpoint, what we will do, the poll function like this. Suppose in this video ability, this will be of the caregiver type. And this again will be popping me get for the first four. We get four because it can be storing the last value. So since phosphate It's eight, then we pushed towards affordability, the last element on the stack, so that will be very close to one and default stops. And the second, second, both this eight will be formed and we will be simply the operator which we got. We will be simply only one function that forward how we will be operating in this case, we will be simply calling function and we will be checking if it is a plus operator. Then in that case we will be working like this. B plus B simply be is. So first of all, first of all, we have formed this and we have collected in this game. And then second time you are hoping we are connected and you can see the order we're doing. B plus a b are not doing APSP. This is what we will be evaluating in this manner. Again, that whatever result you get from the breathless big beaker. So you know that from the plus, what we are getting actually, from P plus a, what we are getting, we are getting 12. This resultant will be, will be simply again, pushing it onto the stack. What we will do and getting me to scan the other symbol, next symbols, but the next thing is true. It isn't. It is fail-stop all the Nazi moles, they are simply pushing this operate it is an operand, so we will be pushing it onto the stack. Again, we will address embolus to be again, pushing it on top of the stack. Once we get the raise two, then we will simply pop elements from the stack. So this is how we will be working and we will be simply checking for this race to the function. That function, let me be, we can simply have given you a switch case. What we will do this sport plus you have seen me God Ford here. We got here. How are you going to be operating? So first of all, we will be operating as b plus a. You can see the order first, second year pork order. You can see we are doing this explanation b plus a. So we will give the switch case better. If it is plus, then we will be simply evaluating in this order. We will use this operator plus. And really it in the order will be plus a n naught a plus B. So the order is very much important because the next time that we see how to convert the infix puppet effect. And that is the, we have to also check what will be the order. In that case, this order you have this orbit is very much important. That is b plus a. So that's what we will write this. Which case if it is a plus operator, if it is a race to operator, if it is a minus, if it is the region. So for all these operators, we will be doing the switch case and we will be writing the expression in this industry. So you can see here, let us come to the function and see now. So first of all, in this evaluate and score post, you can see we have taken a for-loop via picking up on you because I'm scanning the wolf was fixed expression one-by-one, each symbol from left to right. I'm checking if this condition I have, I'm taking in this region, that is if it is done on blood from 0 to nine, then that is if it is an operand and we are simply calling which function we will simply push that operate on to the stack. If it is not 0 to nine, if it is not a number, if it is not an open, then it will go to the else condition. And that means it is an operator. It will be two times so far the first ball I'm collecting in the variability. What does this variability? It is older. You can see here, this is why I'm an longer requisite this end. We are taking this expression as a number. We're not making the expression as a character. So therefore I'm using a long int, read an extra unmuting long in folder, or declare, declaring the variable that is eNB depth. I'm considering this expression which is having the numbers and not the characters. So this is the, this you have to keep in mind. We are taking the expression that was fixed expression as a number. If you're not taking in terms of this, we are substituting the values for the expression. Before we are collecting in this a and the next block we are connecting into this variable. Then we are using the suitcase, this switch cases for what? If you are checking whether this then we are copying. That is for this. You've caught up on, you are storing this. When you get gave you guys some people watching on this condition will be satisfied. That is this. If condition will be satisfied. If it is an operand, if this folder, it will again, condition will be satisfied. The login push it next time. Unless involved will rescan. I evaluate for the, for the for loop at this board will be scanned. It is an operator. Condition is not being satisfied. Else condition in which satisfiability popping from the stack, whatever we have 42 times involved in this. That is for this operator will see whether it is a plus operator, debit autism minus ordinary type. It is a multiplication. If it is a division, if it wasn't age, like if it is array student. This we simply evaluate in this manner. You can see the order. Then is that in-between this operator, whatever the scanning symbol. Again, you can see because I minus this, again for the order is b minus a. If it is a multiplication owners bead. So you have to send the result which will get you use stored in the stem, so on. So it is all of the datatype elongated because it dissolves and the data type and you have your doing putting the brakes everywhere. And finally the resultant, the result. Each time the result you get, suppose you bought some reason than the biggest happens and then you are pushing this stem. Whatever the result you get from B plus a result, the net is this one, B plus EUS looting. In this step, you will be pushing onto the stack. Again, you will be scanning. And the other symbol on this end often you will simply push it onto the stack. If it is not an open-ended is an operator, you will pump the values we have already seen in on the whiteboard. How does evaluation happens? Finally, when you are scanning the whole false fixed exploration, evaluating the whole expression, when you are done with that, in that is you will read your in your stack. That will be uneven venues we witness and bleed benefit from the stack and you will be. In this result. So that will be also on the datatype long because it will be an integer. And finally, you will be returning this value that is already valuable on, but after the evaluation of the expression and you will be returning that value that is returned result. That it will be returned from where it was called. It was called from this position, from this line in the main function. So when we are collecting in this file, you can see I've taken this long int, and finally, I'm printing this value. You can see here. I've hoped so you understood how to convert in fixed tool was fixed as well as how to evaluate it was fixed expression. And we are done with checking with the priorities when, if it is an operator, when we convert infix to postfix to all these fancy, we have seen what should be the result of this final exploration. We are taking this in fixed expedition in the example. That is all should be first of all, you know that first of all, how, what will be result internal event when we just cross-check so we know that this parenthesis me just manually, let me just manually calculate the result of this explanation. We are not using the stack just to find out what will be the result so that whatever coding we are done, we are doing, we are getting the same result or not. So we know that this will be executed manually. Evaluate this exploration without using this tax or this, you know that this what will happen, eight plus or minus two days to six divided by two plus. First of all, this should be evaluated if evaluated manually superpose. We know this will not evaluate cost because supposed to restrict who is debits having higher priority. So today's include having high purity. This will be evaluated and not this one. So therefore, it will be in this manner. We are doing manually this order BY you can do this to find out whatever result we get from this. Villi should be the same. What we have done that with the coding. So in the boarding the logic is different. You're not scanning the world expression again and again. It isn't easy and we're working from an old speaks using the stack so that we don't need to do all these steps. I am doing this just to find out the results of whatever. When I execute my program and we execute our program, then the results should be the same. Just to cross check what is the result of this illustration I'm showing this manner. You can see here like this. Then again, the same thing. It will be like this. That is, it will be plus, so it will be simply six here. Then finally, finally it will be simply, you can see here, this will be, then, if you see here, this will be so if BCC here, this will be minus plus six. So therefore it will be simply this expression will be. Finally, it should give me the explanation is 17. Just to check what would be the result of this, you can already calculate the result of this. You will get the result of this as like the 17th, the program that you are done, we know that what we are doing, we are converting this in fixed too, is fixed. We are converting infix to postfix. We are converting in this manner in which you can see parentheses or not. And we will be scanning from left to right in one group and we'll be evaluating it. So therefore it is very faster because we are using the stack. That is the reason you are converting the infix to postfix using the stack because it is very efficient. And in one more, we can evaluate the resulting expression. You can see how you have seen how he ever been evaluated. So the result should be the same. That is, when we execute, we should get 17 the same as we have seen how we got. Let us simply execute. We know how to execute in this view that to get the loss value, you can simply see here, I have checked this loss. When you check this loss in the view, then you will get the, you already see the belonged here, whatever better you are getting an a compilation error. Or if there is no error, then you are pretty good for running the same program. If it is not in the data and whoever than Mellon good, you can simply execute your program. You can send these on your program. Therefore, I will simply just a second. I'm simply combine the current file. You can see here, there are no arrows, so that means I can simply run it. But then you can see here in the building and run the font sizes are very large and very small. It is ridden infix to postfix and evaluation of the postfix and enter in fixed this message I have written in the main function, or you can see 7. Infix to Prefix Conversion & Evaluation of Prefix Expression C & C++ Session6: Hi, welcome to the new session for writing the program and executing bought infects do prefix button budget, as well as evaluation of the same resultant prefix expression which we obtain after conversion on this fixture prefix will be writing program and executing in C as well as C plus plus Windows operating system. I hope so you have gone through the earlier session where we have seen we have written the program as well as executed for infection was fixed margin and we have evaluated for the postfix exploration. Let us see the scene or the prefix expression two, that is conversion of been fixed to produce an evaluation of the scene. Here we will be considering the same, in fact, expedition which I have on sort of taken on the audio session icon, wanting to postfix explanations that are given to understand it rolled back but the same fixed explanation. And as you know that we have covered in a couple of sessions, what is the fixed expedition? We know what that infects expedition depending upon the position of the operator, it will decide that an expression is in face, but it fixed or pore space in fixed, meaning the operators. It is present in between the operand. So this is an infix expression. You can see here. In-between 84, raised stupor is busy with me to divide as president between 62 and so on. We know the object and by the ICA in fixed to perfect. The reason behind is that we know that if there is an infix expression, then evaluation takes lots of it is very time consuming. That is a reason via the in, from infects pseudo, prefix and login using the stack because tax plays an important role in order to evaluate the expression very weakly positive built on what we'll be working to the prefix using the, if, we can convert infix to prefix. And in this prefix, the prefix which we'll be getting, there will be no parentheses and the priorities will be arranged in the sequence order. And so then we scan it, didn't want to go, we don't need to. It will lead us. It'll be on the board. You don't need to do the repeated scans. This is how it avoids the time consumption. And that is the reason we are using the stack here. So these are the applications which we have already seen for the stack. That is, application is that in order to evaluate the expression many fats that are different types of convergence in Pix2Pix, in fixed reports, fixed. So these are the application on the stacks and movie. I think the program and executing for this program. So let's move on to the or blocks or for this detector, which is required is very easy. You just need to download the code block and you need to install it. It's freely available. After downloading and installing the woodblock, just open a new project and see, suppose we will be seeing the same pilgrimage, be able to recover on your section on the whiteboard for this fixed IP prefix and the evaluation tool here. Moving to the code block, I have already created a new project or in Fitzgerald prefix can login as well as for the evaluation. So this program will be covering two parts. First of all, milk and what and fixed to prefix. Then we will be evaluating the result entry picks, which so far I stop on the menu create approved day for C plus plus. This is your domain, dot cpp. Suppose we will be seeing the program and executing for the C plus plus. And then you will be also seeing the program and policy language and maybe executing it. So this will end. I will also make the board available to you. So you can simply copy, paste the same piece of code and execute on your answer. It will be better if you do the hands-on so that you will understand the concept more easily. Let's see. This is main.cpp, CPP for them, for this program. Here. First of all, you have to include all include all these header files which are required. So I have included iostream, mad Gore-Tex, all these header files which are required. What are the inputs we have already seen in earlier session inputs which are required first of all, being really required one stack using every. Then you require one another area where we will be storing one by one, we will be storing. This involves storing within the prefix explanation. So therefore, one area is required for the prefix expression, one character is required, and one area, as I told that stack using arrays also required to do. Then finally one more that is required that is fixed to restore the infix expression. The duet is my daddy are in fixed expression. One Eddie for prefix expression. One stack using EDI is required. So all these three things are required. And I don't know. And out of bed or operations which are performed on the stack. We know that which operation pop operation. These are the operations which are required for the stack. And the domino that top will be always deferring to the element on the stack. So to the Gulf wars and human element we just pushed really lost when we bought very first we know this concept, taught this deck right? Now. This is your row C plus plus main.cpp file. First of all, here we know that in case of the glass C plus plus program via be required class, I have created here the class by the name in fixed tool, but he fixed this class and fixture prefix. You can see here I have made on the member functions as public. So this is a public access specifier. I have created the constructor and destructor functions without required. I have declared the oblique, that is the membrane Radiodurans. You can see I had made as brightening this look nice, which I'm going to be using, but in fixed IP prefix. Now, which functions are used in this glass you can see the required push. Yes, we require a ball. Then we require in fixture prolific since we in this function, it will be converting from in fixed expression to prefix expression. So dysfunction is going to do the same. Then the another function that is evil underscore prefix that will be evaluating the result in prefix expression, which we then another function floridly so we know that we have been, what is the help us solve the priority. So whenever there is an operator, then we will be checking the rarity of the incoming symbol with that of the priority of people for the stack. So we know that we have already seen in an earlier session what are the incoming symbols, though, in fixed exploration in between these scanning one-by-one, each character will be your symbols first of all, and that is denoted by scans inborn, that priority we will be checking with that of the priority on top of the stack. We know the top of the stack will be the reason. And even that was pushed. So that will be the top of you will be comparing a backup, somebody without that, if you're on the top of the stack and then it will start different. Sebastopol. Come back to that topic in fixed or false fixed conversion so far tougher. Let's go main.cpp. We have seen now for the class which functions are used just in second in this class? In this class, then once I use this function is empty. We check whether the stack is empty or not because if you are performing the pop operation, we have to first check whether the stack is empty or not. This stack is empty, began odd ball height, then end up. Then. Then see here the private access specifier, what all these member variables are there and that is white underscore space. We will see what is the purpose of that. Then another video, this long instructor. So here we are using the stack and this we will be using. Here you can see here the stack which we'll be using this as a stack using EDI. And therefore you can see I have used this subscript and this is max. So this is a stack using edit exist static edit in which you have to define this ice enter compile-time. What does this max? We have, we have already defined the size of the stack. You can give you compile them. You have to give this size. You can change the size of the stack and an anti cancer. And their top dope is referring to the recent element of the stack. Now one thing is dead in this program important fixed to prefix conversion though stack datatype is on the long and it is not on good data type character. So what might this data type is longer because we have already seen or used session in on the whiteboard that while converting from in fixed to prefix and motion was to fall in this stack, we will be storing operants and we will be not storing up operators. So in case of no infix to postfix conversion, what he was doing, it was totally different. The stack we are storing the operators. In case of the infix to postfix conversion, we are not storing the operators on the stat yesterday, the operands that we're the operands which will be there since we will be u, v will be directly substituting it with numbers, that is with the values and the depth order the numbers out of the datatype longing. And therefore we are taking the stack as long and since we're storing operands on the stack, so I won't tell you what the difference between this fixed to prefix and infix to postfix. If you have gone through the video sessions, couple of sessions. Now. Now just we have seen clause and fixed duplicates. Now, moving on to the main dot CPP file, which is a main image is a mean function on this. Last, whenever you Execute your program. The control comes in this main function and nine by nine instructions, and it'll be, all these instructions will be executed sequentially. And you can see here in the very first line I pick indigo is convicts to store that in fixed explanation. That will be at a detailed guide. And that way you can see the IU is a character and we are writing this up code by audit locating the Adi. And then another area we are using prefects so that it isn't able to be stored in one by one in this prefix adding disability that is ultimately effects of the gun voting in fixed too. So two areas are required. For now. We will display the message CLV. Know that SEO is to display the message on the console in C plus plus. It is saying in fixed to prefix and evaluation of prefix in C plus plus. So this is a C plus plus program is done in the C plus plus. This is just a message on the console. Then we are using logins. You'll see what is the purpose of it and enter the infix expression. We are displaying this message on the console. Once this message comes on the console, user will understand that user this holds true and the infix expression, so you're observing enter the expression. So that is a reason we're calling this getter dysfunction. In this in fixed. Whatever expression the user get on the pencil will be stored in this in fixed adding. Now this infix headache will be the input function that is in fixed to prefix, whatever function we will be defining, releasing the definition of that. But we know that in fixture prefix function, the input an infix expression. So for that meeting, what coding is required for, since this is a class that I've created, then I have to call them fixed to prefix functions of waterfall and create the object, static object of the class and fixture prefix in this manner, this is my object and do not buy this object. I'm calling the function that is in fixed tool perfects. And I'm passing this eddies that is in fixed and prefix. So this fixed cost of one, this is the input expression which using them and not on the console. So this is the constraint, the same fixed expedition, which I have shown. This as a false argument. The second argument, perfect. So this is nothing that will pass as it is. Simply characterize research will be filled in this function. It's there. Let's move on to this and fixed it. We fixed. And let's see what coding is required. That function. In case of the fixed prefix, this approach argument isn't fixed expression and secondary human is your prefix expression. That is the, that is the result in prefix expression. So you can see I've given the name by the RDB. Flip it. I will tell you what. I am given this name, but this will be the result in prefix expression. First of all, we have seen in the earlier session on the whiteboard while converting to infix prefix false. The main thing is that just added move on to the notepad. In fixed expedition, which we will be giving unconstrained the same example which I consider infix to postfix also the same in fixed if we indent on the console. In that case, suppose the following expression either be post or reversing. So what is the logic? We have two volts reverse this whole expression one by one. So that is converting fixed. Fix. The medieval exchanges. What is after reversing in fixed walk even get this last node's got activated possible resistor ordering the edit is one thing is stored in. The last will be stored as a 0 or deletion in. And if you know that at indexing starts from 0 to n minus one, so therefore it will be in this metal. Then second Nasdaq is two we have already seen in the diagram on the whiteboard. It will be stored, it will be stored in a second. It'll be the second element. Then this is going to be the land is two will be minus then two and so on. You can see you and this is the octet reversing. You will get this expedition as like this. So after reversing, even get this expression like this. And now you will be scanning this expression from this as it is, that is from Neptune, right? So we obtain this reverse fixed expression. Now the expression you will be scanning from left to right, so forth. Conversion of in fixed to prefix the input port on your session or conversion part in fixture for speaks me don't require to reverse this string. Vr simply scanning from left to right. But in case of conversion are fixed with lipids, we have to reverse the string and that is corrected edit. Once we, once we get the reverse character array, then we will be scanning from left to right. Then the process will be the same, but we have to reverse it post. Coming to the program here, we are declaring some variables here. You'll see what is the purpose of these variables. You can see here, I have taken one character array. Even see what are the use of these videos? Follow along even see now this one. Let us character array. You have seen the IDB IN, infects. Why I'm big in this video, but as I told you that in fixed which is there, I will be reversing it. So therefore, I have named by this name this character areas RDB vivo in fixed. Restore that the reverse of this ambiguous in this edit, this prefix actually, what it is. We will see the old result B2B obtain. You've been stored here one-by-one, this prefix, right after converting. So after applying the list one by one, we will be storing symbols in the prefix expression. So therefore we are taking a character array. Now, this logic is used to reverse in fixed, which is the code in order to reverse it as an amino deck, how to reverse a characteristic, we have to suggest the for-loop from M2 0 position and the act since we know that the end element will be the 0th element. And therefore you can see I've taken the for loop. I've started from streamlined in fixed minus one. So therefore, I indexing will be the index of the last element. And then we have to decrement. You can see here I am decrementing one-by-one I minus, minus the impacts. And what is this? K is equal to 0 V. I've seen that since we will be storing this new Adi, that is RGB pixel, therefore it's indexing starts from 0 to n minus one. So the last element of this stored this 0th position, this new Eddie. And there could be a ticking kids equal to 0 while this IDB infects. And then we also incrementing for this case. So because we will be stalling button by one last element to the 0 position. Therefore, we are taking iron key. I will be referring to this infix actually that started from last. So therefore, I is equal to string in fixed minus one and k is equal to 0 because we are restoring in this new Eddie, therefore, index millimeter, the zeros in restoring and the 0th index of the new Eddie. So therefore you can see here is fixed, that is I. So I will be since IV as typing from the last index of the array that is of the impacts that this last element will be stored as a 0 and human on this new area. So therefore k is equal to 0 initially. Then after the execution of this line, then the I gets decremented by one and that gets incremented by one. Now I, therefore, this condition will be checked. That is, whether I is greater or equal to 0, it will be good it does. So we have to do this in reverse from n to 0 position. Therefore, this condition is checked in IV just to 0. We have to keep on storing one-by-one the element in the new element in x. So this is how we are doing this outreach and is used to reverse the infix expression. As I told you that. First we will be reversing the infix expression after we obtain, after this ONE execution began the reverse string and we will be appending a null character in this result in diverse in fixed expression. Now I've just printed it the reverse and fixed expression. So I should get as like this. Now we will be scanning from left to right. And therefore we are taking the four, we are taking the for-loop in which will be the switch case by such case, because there are different fluids, which we have to check. If we get a closing parenthesis, we get an opening parenthesis. All these rules are that if we get the operators that is like this, if we geta operands, operands is default. So therefore the search space is must for the same or to check what all the roots. And this will be scanning this and it's all done in divorce and fixed exploration from left-to-right definitely started. The formula is equal to 0 I less than string length of this. We've always infix expression. So now we have seen for the early apart in fixed reports, fixed that. Now we will see what we have already seen the also the words on the whiteboard, but in Pix2Pix, what are the ruins? Let me device quickly washed up on the lease out that this is your reverse in fixed explanation, which you have to scan from this left to right. So first of all, when you get the closing parenthesis for conversion of in fixed to clip x, in that case, you have present bleed. Bush this closing parenthesis onto the stack. We have seen that the ending value most converting into we have seen that when he was unloading. We have seen that when he was in fixed reports fix Monday god, opening parenthesis, then we're supposed to push, don't need better. This is onto the stack. In the case of conversion of Pix2Pix, it is important we have then we find the closing parenthesis, then we have to push it onto the staff. Definitely have seen also on the whiteboard. Now, this is the best rule when we get, then another one is that when we get this opening pattern persists, then we have to put all the elements from the stack month by month till you get this losing bet instances. This is also the depth of the infix to postfix in fixed reports fits what he was having. Men began closing parenthesis then we supposed to fall upon the elements till we get to opening material. Men began opening parenthesis. Then you have to walk all the elements till we get to round brackets. So this is the difference in case of fixed IP prefix. No. Then we get, when we get the numbers down here to simply push it. Therefore, we have seen so many get, we get the operands S2 as that. Then we have presumably stored it in the body fixed expression that is in the prefix corrected. When we get an operator like we get this race to this minus or this plus this plus division on these operators. Then we have to deal with that on the stack. Then the picture comes on the priority. We have to check the priority of these scans and one of the priority of the top of the stack. First of all, what are those? What are those? For the operators? If we notice first start one by one. Which case then you'll understand. So first of all, we have already seen in audio session on the whiteboard, we have already under. I've already explained you the folding detail, but let's repeat again and shot. Let's move quickly. Because the main objective of this session is true execute potential the output. So animal and so I have shared the same food so that you can copy, paste the same code and run it at your end. For C plus plus N4, C9, which this for-loop which is dead as it will be up to scan this derivatives in fixed explanation. I'm scanning this one by one. I have taken the for loop. Now I will be storing this reverse that decided to be stored in each character, each character. And I will be checking. And Isaac way to 0. That gives you in fixed subscript 0, the zeros element, that is this one. That is this one going to be like this. This is zeros element of this diverse in fixed expedition one-by-one each connected, ideally scanning and seeing. Which rule does it fixed. So therefore, you can see here I've collected in this symbol. This involved is the datatype character. You can see the symbolism of the datatype character. Now. Now I am proceeding with, I'm proceeding with this things. So I have put the if condition. I have already mentioned. The purpose of this whitespace dysfunction is there. So what this function will be doing and show you this is a function. So you can see I've mentioned the name of the class and thus scope resolution. And because I'm defining the function outside the class, therefore I'm using this class named scope resolution and name of the function I'm defining. So first of all, you can see here the main objective of dysfunction is debt. If you're seeing born because of blank or if it is static, then it will return 1. First of all, we have to ignore me up to not scan. Suppose you get like this, after divorcing expression, you'll get like this. Suppose that have given like this. After divorcing, you will get like this. You can see when you added this lemon, No problem, when you are at this. This is a space where you have to ignore the space you don't need. There are no rules for this face so that you have written the function, right? And the squarespace.com and check if it is a space or a tab, then simply we will return one. And then in this if condition is satisfied. So you can see here this if condition is this IP can be cumin not satisfied and even not check for the roots. We have to avoid impacts of whitespace and proceed with the next Next step element of this EDI. Now, let us think so. Therefore we have put the switch case in this if condition now, switch symbol and the symbol, we'll see if it is, if it is a closing parenthesis, we know we're supposed to simply push it onto the stack. How, what is the function of the push? In case of the stack, we have seen just since this weekly moving, this repeating what is false, which means simply it is an operation on the stack. Whatever symbol you push here, we have to force people pushing you up to see whether your stack is full or not. If the top is greater than max, then you have to give them a seat. You cannot. A stack overflow you that you are not and you cannot push in that case, so you have to simply exit. The stack is not this condition is not satisfied. That is, there is some space on the stack. Then in that case you will simply the stack, you install the symbol and then this top will be incremented. Know, moving through the function again, then break. So if you get done, in other cases, 40, Uganda opening parenthesis, then you have to poke, you get this closing parenthesis. So therefore, this condition is that you have to keep on pumping and you have to keep on scrolling in this prefix, exploration and drilling human p plus plus. So this is in this App Store, do formed evaluates to false menu and the bendy false media we're collecting in the formula when you, what is this positive value? It is the datatype get. We are storing that and then that formed the value stored in the prefix one-by-one. So this will be popped and stored in the prefix. It'll be don't get this opening paren. Once we get opening parenthesis, this condition cannot be satisfied. And then I would say this while loop and this break is applied. Now other rules are for the operators. Once we get less operated on, we get minus multiplication division mode. You can see there are no breaks directly. This instructions are there. In strong, the same instruction will be executed for all these cases instead are no brain brachial. Now, what does this instruction we'll, first off, what does it, what does it roof on the operators? They don't know about? Because in case of in fixed with prefix conversion, then we get random priority. Although let me come here when the polarity, suppose this is Amanda D. Let's consider if you get any operator like raise to minus, plus the priority of the scans and wondered is of this. Suppose this is your scans symbols. If this reality, if it is less than that of the pop from the stack, then it is lesser than the priority of the top of the stack. So in your stack that is, the top will be the last element which was poached. The end that will be the operator on me. So you have to check the priority of the scans and born that is this one. Here would actually then with that of the rarity of the top of this Internet. So if this parity of the scan symbol is lesser than the priority of the top of the stack in this time. And I divided it over the top of this diagram. We are simply supposed to fall the top of the stack and we're supposed to simply store it in the prefix expression. That another rule is this priority on D. I'm just mentioning you that ruins for the for the operators, the priority, the scans symbol, it is greater or equal to priority of the top of the stack. And that gives me up to simply push the scans and staff. So these are the roots. You can see this is if you match with that of the conversion of infix to postfix, this rule, if the priority of the scans and ball is lesser than the priority or the top of the stack to save. If you check with that of the priority. Or again, if you want to check this rule that if the clarity of their scans we want is greater than that of the priority of the top of the stack. That is also somebody as opposed to push them scans and one onto the stack. But that rule is different than the priority of this scans and bull is equal to the priority of the top of the stack, then we are not supposed to fall. In that in the case of in fixed reports, fixed, recheck the associativity, then it might go. But here we will simply push it scans and ball onto the stack. This is a dependence in the control in disarray. So therefore you can see here this foundation which is first of all, checking whether in this while condition, while loop is there, what does this while loop forward? So first of all, this is for us to check whether your stack is empty or not because we'll perform the pop operation, we will be checking post in the clarity of these scans symbol is necessary and the polarity of the top of the stack them, it'll be popping. So before PEPFAR being, you have to always check whether your stack is empty or not. That is, if the stack is empty, there is nothing president, you cannot go. What is this empty function? Let me show you your discord is I had made available to you. You can simply copy paste it. You can access it. What does this empty is empty. We'll check if your topic is minus one, then it will return one else, return 0. Then this loop is for that, that is repost. You will check whether your stack is empty or not. Then you will check your clarity of the scans symbol if it is necessary. If it is lesser than that of the rarity of this tack on the top. This disparity function we have already seen in earlier session, it will simply return the priority of the operators. So if this scan symbol, suppose if it is plus and if you are a stack of their top is multiplication, we know that the plus operator is having the lesser priority as compared to the multiplication. That would show you the clarity function. So you're going to understand it more weakly. I've already shown you multiple times. Just to repeat. This is your priority function. Here you can see again there is a switch case, so there are different varieties. If it is a closing parenthesis, I'm returning 0 because we have seen the advent that there is a closing parenthesis. We have present people shift onto the stack or plus, minus and returning one. Multiplication division was indeed I am returning to for at least two and you're done in three. You can see here the cities. This one is having the highest priority as compared to the other operators. So you can see here, this is just a different return values. Now, if you see here that in this case what will happen? Just, let me show you check the priority here. You check the priority or the priority of the symbol. It is lesser than the polarity of the stop of this tag here. As I told you, if it is plus and this is a multiplication, this condition will be satisfied and this tag will be formed and will be stored in this effects. This you have to keep on doing till you find the priority of the scans in bold lesser than the priority of the top of the stack, minus the polarity of those scans and bone is greater than the validity of the top of the stack, then this condition will not be satisfied. We come out of the loop, your scans in blood that will be pushed onto the stack. I hope so. You understood. One more thing is, what about when you are priority of this in my duplication, if your scans and bond is multiplication and the top of the stack is plus, in that case, first of all, this condition will not be satisfied. This is multiplication, this is multiplication. This is top of the stack AS because this multiplication is having higher priority two, this condition, I'm just seeing different, different rules. What will happen in the priority changes, whether what this condition will be satisfied or not. So there is a reason I'm just mentioning different routes. So if the scans and maleness multiplication and at the top of the stack is plus, then this condition will not be satisfied and this while loop will not be executed that it leads you on SAP, you are pushing the scans in boy, onto the top of the stack. You can understand where the priority of these scans and bold is greater or if it is equal to the priority of the top of this, then only this condition by loop will be executed. And therefore you've been recently pushing the scans involved onto the top of the stack. This is what it will happen. So you understood what will happen, then there are different operators. Now the point that is actually in place all the default, what will happen? The pole will be if anyone having these scans in bold as our openings. In that case, you've been simply prints on your result in prefix. Therefore, the default cases for the same to store the if there are often simply store it onto this prefix edit. These are the dependent rules for all these, oh, this parenthesis operators end for the operands. Now finally, when you are done scattering from starting to last position on the divorce and fixed eddie, then this formula comes to n. That is, if you are done scanning this whole expression, that is from the year two and then four loop comes to the end. And that case, nastily have to check it's still if your stack is not empty, you can see if your stack is not empty, then you have to keep on pumping and you have to keep on storing in this prefix. Then finally, after storing all the elements, after popping on the lemon stronger stack, then you simply append it with this null character. The last. Finally, you bought this prefix fixed expression. Now student job is not completed. You are left with one more task. Nowadays, expression which you obtain after applying these rules and after repulping all elements from the stack stuff, that expression that you get. Now again, you have to reverse this expression. The expression which you get after reversing. That will be the final expression, which is the prefix expression. So I hope you want to understanding. If you haven't already seen on the whiteboard or the prefix you bought. Now again, I have used this for you to simply reverse this fixed display, fixed expression. So you can see again, since actually the most I've been stored the last element of this trip, etc, to the 0th position. So I have taken the new array by the name reverse prefix. You can see here this added, which is this expression which I got better, I will be storing. I'm not taking the new array. Yeah, I'm storing in this edit, so this is in fixed to flip x I call from the main function by the object in fixed expression. And I went fast, one added with the one which I was not filling in from the main function. This adding I'm filling this reverse, be fixed. I'm feeling I've been restoring here. This is the one restored in this reverse. So the final expression which I get I will be storing in this reverse perfect. So you and I will see you understanding what I'm saying, this final prefix. Because after reversing this prefix, then the result in which I get that will be the one that we will finally reversed. And so that is an easy filling in this argument, which then use us as a character array that was not filled audio. I hope so You bought funds, this is done, this is reversed. Find a new gets reversed and fixed expedition after this for-loop and then you are appending the null character. This is the final reverse prefix expression. Let's move on to the main function. Better. Thrombectomy was calling it. Now we have seen how to convert fixed or prefix me gotta result in prefix expression. Now we're simply printing. This is a main function. Here. We're just calling this functioning fixture pubic. So we got final conversion stored in this prefix. We have simply printing. And then finally we have to evaluate this prefix expression. So again, by this object into P, I'm simply calling the function eval underscore prefix. And this prefix, which is that there is a prefix which I got you. Now, I will simply call this function, and I'll pass this prefix in this function. So what is the definition? Let's see here how to evaluate the script X expression. This is the evaluation. So this is, you see here, this slight difference in this evaluation on most everything is seen as compared to the evaluation on the postfix, but little bit differences there. Let's see that you'll understand it more than Z because on most stone, same as compared to that of the evaluation on the postfix. Like Little Italy difference. This evaluation of prefix wherein the input will be the prefix expression which we obtain. Here I'm simply displaying the message evaluation of prefix. I'm taking these variables, we can see what is the user on that. Now, if you see here, I'm taking the for-loop beginning. What I will do the resultant. Let me move on here. First of all, this is prefix expression which we obtain. I've been showing you when we execute the program. So this is a prefix expression which means InDesign passing in this function that is irrelevant the school prefect so as to evaluate this expression. Now, what will happen in this case? So if you see here, I haven't taken the bottom loop, I will be scanning, but I will be scanning from big name. So if you compare the data of the evaluation on the postfix expression for earlier. If you've lost evaluating from beginning to end. In case of the evaluation of the prefix, we are not evaluating from beginning to end. We're evaluating from big name. And that would require knew I had started from Greenland trip x minus one. Then it will be you can see here we