Data Structures Lecture Notes PDF

Data Structures IT623 Dr. Manish Khare Lecture – 7,8 Stacks Slide 2  Stack  A stack is an Abstract Data Type (ADT), commonly used in most programming languages. It is named stack as it behaves like a real...

Data Structures IT623 Dr. Manish Khare Lecture – 7,8 Stacks Slide 2  Stack  A stack is an Abstract Data Type (ADT), commonly used in most programming languages. It is named stack as it behaves like a real-world stack, for example – a deck of cards or a pile of plates, etc.  A real-world stack allows operations at one end only. For example, we can place or remove a card or plate from the top of the stack only. Likewise, Stack ADT allows all data operations at one end only. At any given time, we can only access the top element of a stack.  This feature makes it LIFO data structure. LIFO stands for Last-in-first-out. Here, the element which is placed (inserted or added) last, is accessed first. In stack terminology, insertion operation is called PUSH operation and removal operation is called POP operation. Slide 3  Stack Representation  The following diagram depicts a stack and its operations − Slide 4 Stack Representation Slide 5  Basic Operations on Stack  Stack operations may involve initializing the stack, using it and then de-initializing it. Apart from these basic stuffs, a stack is used for the following two primary operations − push() − Pushing (storing) an element on the stack. pop() − Removing (accessing) an element from the stack.  To use a stack efficiently, we need to check the status of stack as well. For the same purpose, the following functionality is added to stacks − peek() − get the top data element of the stack, without removing it. isFull() − check if stack is full. isEmpty() − check if stack is empty.  At all times, we maintain a pointer to the last PUSHed data on the stack. As this pointer always represents the top of the stack, hence named top. The top pointer provides top value of the stack without actually removing it. Slide 6  Stack Implementation  Stack data structure can be implement in two ways. They are as follows... Using Array Using Linked List  When stack is implemented using array, that stack can organize only limited number of elements. When stack is implemented using linked list, that stack can organize unlimited number of elements. Slide 7 Stack Implementation using Array Slide 8  Stack Operations using Array  A stack can be implemented using array as follows... Before implementing actual operations, first follow the below steps to create an empty stack.  Step 1: Include all the header files which are used in the program and define a constant 'SIZE' with specific value.  Step 2: Declare all the functions used in stack implementation.  Step 3: Create a one dimensional array with fixed size (int stack[SIZE])  Step 4: Define a integer variable 'top' and initialize with '-1'. (int top = -1)  Step 5: In main method display menu with list of operations and make suitable function calls to perform operation selected by the user on the stack. Slide 9  peek() operation  Algorithm of peek() function − begin procedure peek return stack[top] end procedure  Implementation of peek() function in C/C++ programming language int peek() { return stack[top]; } Slide 10  isfull() operation  Algorithm of isfull() function − begin procedure isfull if top equals to MAXSIZE return true else return false endif end procedure Slide 11  isfull() operation  Implementation of isfull() function in C/C++ programming language bool isfull() { if(top == MAXSIZE) return true; else return false; } Slide 12  isempty() operation  Algorithm of isempty() function − begin procedure isempty() if top less than 1 return true else return false endif end procedure Slide 13  isempty() operation  Implementation of isempty() function in C programming language is slightly different. We initialize top at -1, as the index in array starts from 0. So we check if the top is below zero or -1 to determine if the stack is empty. bool isempty() { if(top == -1) return true; else return false; } Slide 14  Push Operation  In a stack, push() is a function used to insert an element into the stack. In a stack, the new element is always inserted at top position. Push function takes one integer value as parameter and inserts that value into the stack. We can use the following steps to push an element on to the stack... Step 1: Check whether stack is FULL. (top == SIZE-1) Step 2: If it is FULL, then display "Stack is FULL!!! Insertion is not possible!!!" and terminate the function. Step 3: If it is NOT FULL, then increment top value by one (top++) and set stack[top] to value (stack[top] = value). Slide 15 Push Operation Slide 16  Algorithm for Push() operation begin procedure push: stack, data if stack is full return null endif top ← top + 1 stack[top] ← data end procedure Slide 17  Implementation of push() in C/C++ void push(int data) { if(!isFull()) { top = top + 1; stack[top] = data; } else { printf("Could not insert data, Stack is full.\n"); } } Slide 18  Pop Operation  In a stack, pop() is a function used to delete an element from the stack. In a stack, the element is always deleted from top position. Pop function does not take any value as parameter. We can use the following steps to pop an element from the stack...  Step 1: Check whether stack is EMPTY. (top == -1)  Step 2: If it is EMPTY, then display "Stack is EMPTY!!! Deletion is not possible!!!" and terminate the function.  Step 3: If it is NOT EMPTY, then delete stack[top] and decrement top value by one (top--). Slide 19 Slide 20  Algorithm for Pop() operation begin procedure pop: stack if stack is empty return null endif data ← stack[top] top ← top - 1 return data end procedure Slide 21  Implementation of pop() in C/C++ int pop(int data) { if(!isempty()) { data = stack[top]; top = top - 1; return data; } else { printf("Could not retrieve data, Stack is empty.\n"); } } Slide 22  Display operation  We can use the following steps to display the elements of a stack...  Step 1: Check whether stack is EMPTY. (top == -1)  Step 2: If it is EMPTY, then display "Stack is EMPTY!!!" and terminate the function.  Step 3: If it is NOT EMPTY, then define a variable 'i' and initialize with top. Display stack[i] value and decrement i value by one (i--).  Step 3: Repeat above step until i value becomes '0'. Slide 23  Implementation of display() in C/C++ void display(){ if(top == -1) printf("\nStack is Empty!!!"); else{ int i; printf("\nStack elements are:\n"); for(i=top; i>=0; i--) printf("%d\n",stack[i]); } } Slide 24  Program on Stack using array Slide 25 Stack Implementation using Linked List Slide 26  The major problem with the stack implemented using array is, it works only for fixed number of data values. That means the amount of data must be specified at the beginning of the implementation itself. Stack implemented using array is not suitable, when we don't know the size of data which we are going to use.  A stack data structure can be implemented by using linked list data structure. The stack implemented using linked list can work for unlimited number of values. That means, stack implemented using linked list works for variable size of data. So, there is no need to fix the size at the beginning of the implementation. The Stack implemented using linked list can organize as many data values as we want.  In linked list implementation of a stack, every new element is inserted as 'top' element. That means every newly inserted element is pointed by 'top'. Whenever we want to remove an element from the stack, simply remove the node which is pointed by 'top' by moving 'top' to its next node in the list. The next field of the first element must be always NULL. Slide 27  In above example, the last inserted node is 99 and the first inserted node is 25. The order of elements inserted is 25, 32,50 and 99. Slide 28  Stack Operations using Linked List  To implement stack using linked list, we need to set the following things before implementing actual operations.  Step 1: Include all the header files which are used in the program. And declare all the user defined functions.  Step 2: Define a 'Node' structure with two members data and next.  Step 3: Define a Node pointer 'top' and set it to NULL.  Step 4: Implement the main method by displaying Menu with list of operations and make suitable function calls in the main method. Slide 29  Push Operation  We can use the following steps to insert a new node into the stack... Step 1: Create a newNode with given value. Step 2: Check whether stack is Empty (top == NULL) Step 3: If it is Empty, then set newNode → next = NULL. Step 4: If it is Not Empty, then set newNode → next = top. Step 5: Finally, set top = newNode. Slide 30  Algorithm for Push() operation Slide 31  Pop Operation  We can use the following steps to delete a node from the stack... Step 1: Check whether stack is Empty (top == NULL). Step 2: If it is Empty, then display "Stack is Empty!!! Deletion is not possible!!!" and terminate the function Step 3: If it is Not Empty, then define a Node pointer 'temp' and set it to 'top'. Step 4: Then set 'top = top → next'. Step 7: Finally, delete 'temp' (free(temp)). Slide 32  Algorithm for Pop() operation Slide 33  Display operation  We can use the following steps to display the elements (nodes) of a stack... Step 1: Check whether stack is Empty (top == NULL). Step 2: If it is Empty, then display 'Stack is Empty!!!' and terminate the function. Step 3: If it is Not Empty, then define a Node pointer 'temp' and initialize with top. Step 4: Display 'temp → data --->' and move it to the next node. Repeat the same until temp reaches to the first node in the stack (temp → next != NULL). Step 4: Finally! Display 'temp → data ---> NULL'. Slide 34  Program on Stack using linked list Slide 35  Multiple Stack  While implementing a stack using an array, we had seen that the size of the array must be known in advance. If the stack is allocated less space, then frequent OVERFLOW conditions will be encountered.  To deal with this problem, the code will have to be modified to reallocate more space for the array. In case we allocate a large amount of space for the stack, it may result in sheer wastage of memory.  Thus, there lies a trade-off between the frequency of overflows and the space allocated.  So, a better solution to deal with this problem is to have multiple stacks or to have more than one stack in the same array of sufficient size. Slide 36  An array STACK[n] is used to represent two stacks, Stack A and Stack B. The value of n is such that the combined size of both the stacks will never exceed n.  While operating on these stacks, it is important to note one thing—Stack A will grow from left to right, whereas Stack B will grow from right to left at the same time.  Program on multiple Stack. Slide 37 Applications of Stacks Slide 38  Applications of Stacks  Stacks can be used on many applications. Conversion of an infix expression into a postfix expression Evaluation of a postfix expression Conversion of an infix expression into a prefix expression Evaluation of a prefix expression Reversing a list Parentheses checker Recursion Tower of Hanoi Slide 39 Expressions Slide 40  Expressions  In any programming language, if we want to perform any calculation or to frame a condition etc., we use a set of symbols to perform the task. These set of symbols makes an expression.  An expression can be defined as follows... An expression is a collection of operators and operands that represents a specific value. Slide 41  Expression Types  Based on the operator position, expressions are divided into THREE types. They are as follows... Infix Expression Postfix Expression Prefix Expression Slide 42  Infix Expression  In infix expression, operator is used in between operands.  The general structure of an Infix expression is as follows... Slide 43  Postfix Expression  In postfix expression, operator is used after operands. We can say that "Operator follows the Operands".  The general structure of Postfix expression is as follows... Slide 44  Prefix Expression  In prefix expression, operator is used before operands. We can say that "Operands follows the Operator".  The general structure of Prefix expression is as follows... Slide 45  Precedence  When an operand is in between two different operators, which operator will take the operand first, is decided by the precedence of an operator over others.  For example −  As multiplication operation has precedence over addition, b * c will be evaluated first. A table of operator precedence is provided later. Slide 46  Precedence  For example, x = 7 + 3 * 2;  Here, x is assigned 13, not 20 because operator * has a higher precedence than +, so it first gets multiplied with 3*2 and then adds into 7. Slide 47  Associativity  Associativity describes the rule where operators with the same precedence appear in an expression.  For example, in expression a + b − c, both + and – have the same precedence, then which part of the expression will be evaluated first, is determined by associativity of those operators.  Here, both + and − are left associative, so the expression will be evaluated as (a + b) − c. Slide 48  Precedence and associativity determines the order of evaluation of an expression.  Following is an operator precedence and associativity table (highest to lowest) Sr. No. Operator Precedence Associativity 1 Exponentiation ^ Highest Right Associative 2 Multiplication ( ∗ ) & Division ( / ) Second Highest Left Associative 3 Addition ( + ) & Subtraction ( − ) Lowest Left Associative  Complete Precedence list Slide 49 Slide 50  The above table shows the default behavior of operators. At any point of time in expression evaluation, the order can be altered by using parenthesis. For example −  In a + b*c, the expression part b*c will be evaluated first, with multiplication as precedence over addition.  But if we use use parenthesis for a + b. Then expression will be evaluated first, like (a + b)*c. Slide 51  Expression Conversion  To convert any Infix expression into Postfix or Prefix expression we can use the following procedure... Find all the operators in the given Infix Expression. Find the order of operators evaluated according to their Operator precedence. Convert each operator into required type of expression (Postfix or Prefix) in the same order. Slide 52  Any expression can be represented using these three different types of expressions.  And we can convert an expression from one form to another form like Infix to Postfix, Infix to Prefix, Prefix to Postfix Prefix to infix Postfix to prefix Postfix to infix Slide 53  Example Consider the following Infix Expression to be converted into Postfix Expression... D=A+B*C  Step 1: The Operators in the given Infix Expression : = , + , *  Step 2: The Order of Operators according to their preference : * , + , =  Step 3: Now, convert the first operator * ----- D = A + B C *  Step 4: Convert the next operator + ----- D = A BC* +  Step 5: Convert the next operator = ----- D ABC*+ = Finally, given Infix Expression is converted into Postfix Expression as follows... DABC*+= Slide 54  Exercise  Convert the following infix expressions into postfix and prefix expressions (A-B)*(C+D) [AB–] * [CD+] AB–CD+* [-AB] * [+CD] *-AB+CD Slide 55  Exercise  Convert the following infix expressions into postfix and prefix expressions (A+B)/(C+D)-(D*E) [AB+] / [CD+] – [DE*] [AB+CD+/] – [DE*] AB+CD+/DE*– [+AB] / [+CD] – [*DE] [/+AB+CD] – [*DE] -/+AB+CD*DE Slide 56  Convert the following infix expressions into prefix expressions. (A + B) * C (+AB)*C *+ABC (A–B) * (C+D) [–AB] * [+CD] *–AB+CD (A + B) / ( C + D) – ( D * E) [+AB] / [+CD] – [*DE] [/+AB+CD] – [*DE] –/+AB+CD*DE Slide 57  Infix to Postfix Conversion  Procedure for Postfix Conversion Scan the Infix string from left to right. Initialize an empty stack. If the scanned character is an operand, add it to the Postfix string. If the scanned character is an operator and if the stack is empty push the character to stack. If the scanned character is an Operator and the stack is not empty, compare the precedence of the character with the element on top of the stack. If top Stack has higher precedence over the scanned character pop the stack else push the scanned character to stack. Repeat this step until the stack is not empty and top Stack has precedence over the character. Repeat 4 and 5 steps till all the characters are scanned. After all characters are scanned, we have to add any character that the stack may have to the Postfix string. If stack is not empty add top Stack to Postfix string and Pop the stack. Repeat this step as long as stack is not empty. Slide 58  Algorithm for Postfix Conversion S:stack while(more tokens) xprefixToInfix(stack) d. ELSE Write temp to output IF stack.top NOT EQUAL to space -->prefixToInfix(stack) Slide 72  *+a-bc/-de+-fgh Expression Stack *+a-bc/-de+-fgh NuLL *+a-bc/-de+-fg "h" "g" *+a-bc/-de+-f "h" "f" *+a-bc/-de+- "g" "h" "f - g" *+a-bc/-de+ "h" *+a-bc/-de "f-g+h" Slide 73  "e" *+a-bc/-d "f-g+h" "d" *+a-bc/- "e" "f-g+h" "d - e" *+a-bc/ "f-g+h" *+a-bc "(d-e)/(f-g+h)" "c" *+a-b "(d-e)/(f-g+h)" Slide 74  "b" *+a- "c" "(d-e)/(f-g+h)" "b-c" *+a "(d-e)/(f-g+h)" "a" *+ "b-c" "(d-e)/(f-g+h)" "a+b-c" * "(d-e)/(f-g+h)" End "(a+b-c)*(d-e)/(f-g+h)" Slide 75  Postfix Expression Evaluation  A postfix expression can be evaluated using the Stack data structure.  To evaluate a postfix expression using Stack data structure we can use the following steps... Read all the symbols one by one from left to right in the given Postfix Expression If the reading symbol is operand, then push it on to the Stack. If the reading symbol is operator (+ , - , * , / etc.,), then perform TWO pop operations and store the two popped oparands in two different variables (operand1 and operand2). Then perform reading symbol operation using operand1 and operand2 and push result back on to the Stack. Finally! perform a pop operation and display the popped value as final result. Slide 76  Consider the infix expression given as 9 – ((3 * 4) + 8) / 4.  The infix expression 9 – ((3 * 4) + 8) / 4 can be written as 9 3 4 * 8 + 4 / – using postfix notation. Slide 77  Prefix Expression Evaluation  The algorithm for evaluating a prefix expression is as follows: Accept a prefix string from the user Repeat until all the characters in the prefix expression have been scanned Scan the prefix expression from right, one character at a time. If the scanned character is an operand, push it on the operand stack If the scanned character is an operator, then (i). Pop two values from the operand stack (ii). Apply the operator on the popped operands (iii). Push the result on the operand stack End Slide 78  For example, consider the prefix expression + – 9 2 7 * 8 / 4 12. Slide 79  Reversing a List A list of numbers can be reversed by reading each number from an array starting from the first index and pushing it on a stack. Once all the numbers have been read, the numbers can be popped one at a time and then stored in the array starting from the first index. Slide 80  Implementing Parentheses Checker  Stacks can be used to check the validity of parentheses in any algebraic expression. For example, an algebraic expression is valid if for every open bracket there is a corresponding closing bracket.  For example, the expression (A+B} is invalid but an expression {A + (B – C)} is valid. Look at the program below which traverses an algebraic expression to check for its validity. Slide 81  Algorithm: 1) Declare a character stack S. 2) Now traverse the expression string exp. a) If the current character is a starting bracket (‘(‘ or ‘{‘ or ‘[‘) then push it to stack. b) If the current character is a closing bracket (‘)’ or ‘}’ or ‘]’) then pop from stack and if the popped character is the matching starting bracket then fine else parenthesis are not balanced. 3) After complete traversal, if there is some starting bracket left in stack then “not balanced” Slide 82  Recursion  A recursive function is defined as a function that calls itself to solve a smaller version of its task until a final call is made which does not require a call to itself.  Since a recursive function repeatedly calls itself, it makes use of the system stack to temporarily store the return address and local variables of the calling function.  Every recursive solution has two major cases. They are Base case, in which the problem is simple enough to be solved directly without making any further calls to the same function. Recursive case, in which first the problem at hand is divided into simpler sub- parts. Second the function calls itself but with sub-parts of the problem obtained in the first step. Third, the result is obtained by combining the solutions of simpler sub-parts.  Therefore, recursion is defining large and complex problems in terms of smaller and more easily solvable problems. In recursive functions, a complex problem is defined in terms of simpler problems and the simplest problem is given explicitly. Slide 83  To understand recursive functions, let us take an example of calculating factorial of a number.  To calculate n!, we multiply the number with factorial of the number that is 1 less than that number.  In other words, n! = n x (n–1)!  Let us say we need to find the value of 5!  5! = 5 x 4 x 3 x 2 x 1 = 120  This can be written as 5! = 5 x 4!, where 4!= 4 x 3!  Therefore, 5! = 5 x 4 x 3!  Similarly, we can also write 5! = 5 x 4 x 3 x 2!  Expanding further 5! = 5 x 4 x 3 x 2 x 1!  We know, 1! = 1 Slide 84  every recursive function must have a base case and a recursive case. For the factorial function Base case is when n = 1, because if n = 1, the result will be 1 as 1! = 1. Recursive case of the factorial function will call itself but with a smaller value of n, this case can be given as factorial(n) = n × factorial (n–1) Slide 85  From the example, let us analyze the steps of a recursive program. Step 1: Specify the base case which will stop the function from making a call to itself. Step 2: Check to see whether the current value being processed matches with the value of the base case. If yes, process and return the value. Step 3: Divide the problem into smaller or simpler sub- problems. Step 4: Call the function from each sub-problem. Step 5: Combine the results of the sub-problems. Step 6: Return the result of the entire problem. Slide 86  Examples of Recursion  Greatest Common Divisor  Finding Exponents  The Fibonacci Series Slide 87  Types of Recursion  Recursion is a technique that breaks a problem into one or more sub-problems that are similar to the original problem. Any recursive function can be characterized based on whether the function calls itself directly or indirectly (direct or indirect recursion), Slide 88  Direct Recursion A function is said to be directly recursive if it explicitly calls itself. For example, consider the code shown in following figure. Here, the function Func() calls itself for all positive values of n, so it is said to be a directly recursive function. Slide 89  Indirect Recursion A function is said to be indirectly recursive if it contains a call to another function which ultimately calls it. Look at the functions given below. These two functions are indirectly recursive as they both call each other. Slide 90  Tower of Hanoi  The tower of Hanoi is one of the main applications of recursion. It says, ‘if you can solve n–1 cases, then you can easily solve the nth case’. Slide 91 Tower of Hanoi Slide 92  Advantage of Recursion  Recursive solutions often tend to be shorter and simpler than non-recursive ones.  Code is clearer and easier to use.  Recursion works similar to the original formula to solve a problem.  Recursion follows a divide and conquer technique to solve problems.  In some (limited) instances, recursion may be more efficient. Slide 93  Advantages/Drawback of recursion  For some programmers and readers, recursion is a difficult concept.  Recursion is implemented using system stack. If the stack space on the system is limited, recursion to a deeper level will be difficult to implement.  Aborting a recursive program in midstream can be a very slow process.  Using a recursive function takes more memory and time to execute as compared to its non-recursive counterpart.  It is difficult to find bugs, particularly while using global variables. Slide 94