Linear Data Structures-Stacks

Introduction of Stack

A stack is a fundamental data structure in computer science that operates in a Last-In-First-Out (LIFO) manner. This means that the last element added to the stack is the first one to be removed. Stacks are used in various applications, such as evaluating expressions, backtracking algorithms, and managing function calls.

Key Characteristics of a Stack

1. LIFO (Last-In-First-Out): The most recently added element is the first to be removed.
2. Operations:
   • Push: Add an element to the top of the stack.
   • Pop: Remove the top element from the stack.
   • Peek (or Top): Return the top element without removing it.
   • isEmpty: Check if the stack is empty.
3. Use Cases:
   • Function Call Management: Stacks are used in programming languages for keeping track of function calls and local variables. The call stack is a stack that stores information about active subroutines.
   • Expression Evaluation: Stacks are used in algorithms for evaluating arithmetic expressions (e.g., infix to postfix conversion).
   • Undo Mechanisms: Stacks are used in applications like text editors to keep track of changes for undo operations.
   • Backtracking Algorithms: Stacks are useful in algorithms that involve backtracking, such as solving mazes or puzzles like the Tower of Hanoi.

Advantages and Disadvantages

Advantages:

• Simple and easy to implement.
• Efficient for LIFO operations.

Disadvantages:

• Limited access to only the top element.
• Not suitable for scenarios where random access is needed.

Stacks are an essential concept in data structures and algorithms, providing a simple yet powerful mechanism for managing data in a controlled manner. Understanding stacks is crucial for problem-solving and software development.

Push operations

The push operation adds a new element to the top of the stack. This is like stacking plates in a cafeteria: when you add a new plate, you place it on top of the stack of plates.

Here are the steps involved in the push operation:

1. Check for Space:
   • Ensure there is room for a new element. In fixed-size stacks, this means checking for overflow (when the stack is full).
2. Add the Element:
   • Place the new element on top of the current stack.
3. Update the Top:
   • Adjust the top pointer or index to reflect the new element’s position.

Example program on push operation

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

// Define the maximum size of the stack
#define MAX_SIZE 5

// Define the stack structure
typedef struct {
    int items[MAX_SIZE];  // Array to store stack elements
    int top;              // Index of the top element
} Stack;

// Function to initialize the stack
void initStack(Stack* s) {
    s->top = -1;  // The stack is initially empty
}

// Function to check if the stack is full
bool isFull(Stack* s) {
    return s->top == MAX_SIZE - 1;  // Returns true if the top is at the last index
}

// Function to check if the stack is empty
bool isEmpty(Stack* s) {
    return s->top == -1;  // Returns true if the top is at the initial position
}

// Function to push an element onto the stack
void push(Stack* s, int item) {
    // Check for overflow
    if (isFull(s)) {
        printf("Stack overflow: Cannot push %d onto the stack.\n", item);
        return;
    }
    
    // Increment the top index and add the item
    s->top++;
    s->items[s->top] = item;
    
    // Print the current state of the stack
    printf("Pushed %d: Current stack: ", item);
    for (int i = 0; i <= s->top; i++) {
        printf("%d ", s->items[i]);
    }
    printf("\n");
}

// Main function to demonstrate the push operation
int main() {
    Stack stack;
    initStack(&stack);  // Initialize the stack
    
    // Push elements onto the stack
    push(&stack, 10);
    push(&stack, 20);
    push(&stack, 30);
    push(&stack, 40);
    push(&stack, 50);
    push(&stack, 60);  // This will cause a stack overflow
    
    return 0;
}

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

#define MAX_SIZE 5

typedef struct {
    int items[MAX_SIZE];  // Array to store stack elements
    int top;              // Index of the top element
} Stack;

void initStack(Stack* s) {
    s->top = -1;  // The stack is initially empty
}

bool isFull(Stack* s) {
    return s->top == MAX_SIZE - 1;  // Returns true if the top is at the last index
}

bool isEmpty(Stack* s) {
    return s->top == -1;  // Returns true if the top is at the initial position
}

void push(Stack* s, int item) {
    // Check for overflow
    if (isFull(s)) {
        printf("Stack overflow: Cannot push %d onto the stack.\n", item);
        return;
    }
    
    // Increment the top index and add the item
    s->top++;
    s->items[s->top] = item;
    
    // Print the current state of the stack
    printf("Pushed %d: Current stack: ", item);
    for (int i = 0; i <= s->top; i++) {
        printf("%d ", s->items[i]);
    }
    printf("\n");
}

int main() {
    Stack stack;
    initStack(&stack);  // Initialize the stack
    
    // Push elements onto the stack
    push(&stack, 10);
    push(&stack, 20);
    push(&stack, 30);
    push(&stack, 40);
    push(&stack, 50);
    push(&stack, 60);  // This will cause a stack overflow
    
    return 0;
}

Pop operation

The pop operation is used to remove the top element of the stack.

Steps to Perform the Pop Operation:

1. Check for Underflow:
   • Before attempting to pop an element, check if the stack is empty by checking if top is -1.
   • If the stack is empty, report an underflow error since there is no element to pop.
2. Access the Top Element:
   • Retrieve the element at the index top in the array, as this is the element to be removed.
3. Update the Top Index:
   • Decrement the top index by 1 to point to the next element in the stack.
   • This effectively removes the element from the stack by no longer keeping track of it.
4. Return the Popped Element:
   • Return the element that was removed for any further use or processing.

Program for Pop operation

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

// Define the maximum size of the stack
#define MAX_SIZE 5

// Define the stack structure
typedef struct {
    int items[MAX_SIZE];  // Array to store stack elements
    int top;              // Index of the top element
} Stack;

// Function to initialize the stack
void initStack(Stack* s) {
    s->top = -1;  // The stack is initially empty
}

// Function to check if the stack is full
bool isFull(Stack* s) {
    return s->top == MAX_SIZE - 1;  // Returns true if the top is at the last index
}

// Function to check if the stack is empty
bool isEmpty(Stack* s) {
    return s->top == -1;  // Returns true if the top is at the initial position
}

// Function to push an element onto the stack
void push(Stack* s, int item) {
    // Check for overflow
    if (isFull(s)) {
        printf("Stack overflow: Cannot push %d onto the stack.\n", item);
        return;
    }
    
    // Increment the top index and add the item
    s->top++;
    s->items[s->top] = item;
    
    // Print the current state of the stack
    printf("Pushed %d: Current stack: ", item);
    for (int i = 0; i <= s->top; i++) {
        printf("%d ", s->items[i]);
    }
    printf("\n");
}

// Function to pop an element from the stack
int pop(Stack* s) {
    // Check for underflow
    if (isEmpty(s)) {
        printf("Stack underflow: Cannot pop from an empty stack.\n");
        return -1;  // Return a sentinel value to indicate failure
    }
    
    // Retrieve the top element and decrement the top index
    int poppedItem = s->items[s->top];
    s->top--;
    
    // Print the current state of the stack
    printf("Popped %d: Current stack: ", poppedItem);
    for (int i = 0; i <= s->top; i++) {
        printf("%d ", s->items[i]);
    }
    printf("\n");
    
    return poppedItem;  // Return the popped element
}

// Main function to demonstrate the pop operation
int main() {
    Stack stack;
    initStack(&stack);  // Initialize the stack
    
    // Push elements onto the stack
    push(&stack, 10);
    push(&stack, 20);
    push(&stack, 30);
    push(&stack, 40);
    push(&stack, 50);
    
    // Pop elements from the stack
    pop(&stack);
    pop(&stack);
    pop(&stack);
    pop(&stack);
    pop(&stack);
    pop(&stack);  // This will cause a stack underflow
    
    return 0;
}

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

#define MAX_SIZE 5

typedef struct {
    int items[MAX_SIZE];  // Array to store stack elements
    int top;              // Index of the top element
} Stack;

void initStack(Stack* s) {
    s->top = -1;  // The stack is initially empty
}

bool isFull(Stack* s) {
    return s->top == MAX_SIZE - 1;  // Returns true if the top is at the last index
}

bool isEmpty(Stack* s) {
    return s->top == -1;  // Returns true if the top is at the initial position
}

void push(Stack* s, int item) {
    // Check for overflow
    if (isFull(s)) {
        printf("Stack overflow: Cannot push %d onto the stack.\n", item);
        return;
    }
    
    // Increment the top index and add the item
    s->top++;
    s->items[s->top] = item;
    
    // Print the current state of the stack
    printf("Pushed %d: Current stack: ", item);
    for (int i = 0; i <= s->top; i++) {
        printf("%d ", s->items[i]);
    }
    printf("\n");
}

int pop(Stack* s) {
    // Check for underflow
    if (isEmpty(s)) {
        printf("Stack underflow: Cannot pop from an empty stack.\n");
        return -1;  // Return a sentinel value to indicate failure
    }
    
    // Retrieve the top element and decrement the top index
    int poppedItem = s->items[s->top];
    s->top--;
    
    // Print the current state of the stack
    printf("Popped %d: Current stack: ", poppedItem);
    for (int i = 0; i <= s->top; i++) {
        printf("%d ", s->items[i]);
    }
    printf("\n");
    
    return poppedItem;  // Return the popped element
}

int main() {
    Stack stack;
    initStack(&stack);  // Initialize the stack
    
    // Push elements onto the stack
    push(&stack, 10);
    push(&stack, 20);
    push(&stack, 30);
    push(&stack, 40);
    push(&stack, 50);
    
    // Pop elements from the stack
    pop(&stack);
    pop(&stack);
    pop(&stack);
    pop(&stack);
    pop(&stack);
    pop(&stack);  // This will cause a stack underflow
    
    return 0;
}

Stack Over flow and Under flow

Overflow in Stacks

Overflow occurs when trying to add (push) an element to a stack that is already full.

1. Array-Based Stack Overflow

In an array-based stack, the size is fixed and predetermined, which means the stack has a maximum capacity defined by the size of the array.

How Overflow Occurs:

• Condition: The overflow condition happens when the top index reaches the maximum size of the array (MAX_SIZE - 1).
• Attempted Operation: A push operation is attempted on a full stack.

Underflow in Stacks

Underflow occurs when trying to remove (pop) an element from a stack that is already empty.

1. Array-Based Stack Underflow

In an array-based stack, underflow occurs when there are no elements left to pop.

How Underflow Occurs:

• Condition: The stack is empty when top is -1.
• Attempted Operation: A pop operation is attempted on an empty stack.

Linked list implementation of a Stack

Implementing a stack using a linked list is a flexible and dynamic approach that allows the stack to grow and shrink as needed without a fixed size limit. Let’s go through a detailed implementation of a stack using a linked list in C, including the basic operations like push, pop, and peek.

To implement a stack using a singly linked list, we leverage the LIFO (Last In, First Out) principle, where operations are performed at the “head” of the list. Here’s how it works:

• Top Pointer: In our stack implementation, the top pointer represents the head of the list. This is where both pushing and popping operations occur.
• Linked List Structure: Each node in the linked list contains:
  • Data Field: Stores the element of the stack.
  • Link Field: Points to the next node in the list.
  • The top pointer points to the first node in the list, and the link field of the last node is NULL.

Advantages of Using a Linked List for Stack Implementation

• Dynamic Size: Unlike arrays, a linked list allows the stack to grow and shrink dynamically without a fixed capacity.
• No Overflow: Since memory is allocated dynamically for each node, stack overflow due to a fixed array size is avoided.

Stack Operations

1. push(): Adds a new element to the top of the stack.
   • Steps:
     1. Initialize a new node.
     2. Set the node’s data field to the new value.
     3. Link this node to the current top node.
     4. Update the top pointer to point to the new node.
2. pop(): Removes and returns the top element of the stack.
   • Steps:
     1. Check if the stack is empty. If it is, report an underflow condition.
     2. Access the top node.
     3. Update the top pointer to the next node in the list.
     4. Free the memory allocated for the old top node.
     5. Return the data from the removed node.
3. peek(): Returns the top element without removing it.
   • Steps:
     1. Check if the stack is empty. If it is, report an underflow condition.
     2. Return the data from the node pointed to by the top pointer.
4. display(): Prints all elements in the stack.
   • Steps:
     1. Start from the top node.
     2. Traverse the list, printing each node’s data until reaching the end of the list.

Program for linked list Implementation of a stack and it’s operation

#include <stdio.h>
#include <stdlib.h>

// Define the Node structure
typedef struct Node {
    int data;           // Data stored in the node
    struct Node* next;  // Pointer to the next node
} Node;

// Define the Stack structure
typedef struct Stack {
    Node* top;          // Pointer to the top node of the stack
} Stack;

// Function to initialize the stack
void initStack(Stack* s) {
    s->top = NULL;  // The stack is initially empty
}

// Function to check if the stack is empty
int isEmpty(Stack* s) {
    return s->top == NULL;
}

// Function to push an element onto the stack
void push(Stack* s, int item) {
    // Step 1: Create a new node
    Node* newNode = (Node*)malloc(sizeof(Node));
    if (!newNode) {
        printf("Memory allocation failed: Cannot push %d onto the stack.\n", item);
        return;
    }
    
    // Step 2: Set the new node's data
    newNode->data = item;
    
    // Step 3: Link the new node to the current top node
    newNode->next = s->top;
    
    // Step 4: Update the top pointer to the new node
    s->top = newNode;
    
    // Print confirmation of the push operation
    printf("Pushed %d onto the stack.\n", item);
}

// Function to pop an element from the stack
int pop(Stack* s) {
    // Check for underflow
    if (isEmpty(s)) {
        printf("Stack underflow: Cannot pop from an empty stack.\n");
        return -1;  // Return a sentinel value to indicate failure
    }
    
    // Access the top node and store its data
    Node* temp = s->top;
    int poppedItem = temp->data;
    
    // Update the top pointer to the next node
    s->top = s->top->next;
    
    // Free the old top node
    free(temp);
    
    // Return the popped item
    printf("Popped %d from the stack.\n", poppedItem);
    return poppedItem;
}

// Function to peek at the top element of the stack
int peek(Stack* s) {
    if (isEmpty(s)) {
        printf("Stack is empty: Cannot peek.\n");
        return -1;  // Return a sentinel value to indicate failure
    }
    
    // Return the data of the top node
    return s->top->data;
}

// Function to display all elements in the stack
void display(Stack* s) {
    if (isEmpty(s)) {
        printf("Stack is empty.\n");
        return;
    }
    
    Node* current = s->top;
    printf("Stack elements:\n");
    while (current != NULL) {
        printf("%d\n", current->data);
        current = current->next;
    }
}

// Main function to demonstrate stack operations
int main() {
    Stack stack;
    initStack(&stack);  // Initialize the stack
    
    // Push elements onto the stack
    push(&stack, 10);
    push(&stack, 20);
    push(&stack, 30);
    push(&stack, 40);
    push(&stack, 50);
    
    // Peek at the top element
    printf("Top element: %d\n", peek(&stack));
    
    // Display all elements in the stack
    display(&stack);
    
    // Pop elements from the stack
    pop(&stack);
    pop(&stack);
    
    // Peek again
    printf("Top element after popping: %d\n", peek(&stack));
    
    // Display remaining elements in the stack
    display(&stack);
    
    return 0;
}

typedef struct Node {
    int data;           // The data stored in the node
    struct Node* next;  // Pointer to the next node in the list
} Node;

typedef struct Stack {
    Node* top;          // Pointer to the top node of the stack
} Stack;

void initStack(Stack* s) {
    s->top = NULL;  // The stack is initially empty
}

int isEmpty(Stack* s) {
    return s->top == NULL;
}

void push(Stack* s, int item) {
    Node* newNode = (Node*)malloc(sizeof(Node));
    if (!newNode) {
        printf("Memory allocation failed: Cannot push %d onto the stack.\n", item);
        return;
    }
    
    newNode->data = item;
    newNode->next = s->top;
    s->top = newNode;
    
    printf("Pushed %d onto the stack.\n", item);
}

int pop(Stack* s) {
    if (isEmpty(s)) {
        printf("Stack underflow: Cannot pop from an empty stack.\n");
        return -1;  // Return a sentinel value to indicate failure
    }
    
    Node* temp = s->top;
    int poppedItem = temp->data;
    
    s->top = s->top->next;
    free(temp);
    
    printf("Popped %d from the stack.\n", poppedItem);
    return poppedItem;
}

int peek(Stack* s) {
    if (isEmpty(s)) {
        printf("Stack is empty: Cannot peek.\n");
        return -1;  // Return a sentinel value to indicate failure
    }
    
    return s->top->data;
}

void display(Stack* s) {
    if (isEmpty(s)) {
        printf("Stack is empty.\n");
        return;
    }
    
    Node* current = s->top;
    printf("Stack elements:\n");
    while (current != NULL) {
        printf("%d\n", current->data);
        current = current->next;
    }
}

int main() {
    Stack stack;
    initStack(&stack);  // Initialize the stack
    
    push(&stack, 10);
    push(&stack, 20);
    push(&stack, 30);
    push(&stack, 40);
    push(&stack, 50);
    
    printf("Top element: %d\n", peek(&stack));
    
    display(&stack);
    
    pop(&stack);
    pop(&stack);
    
    printf("Top element after popping: %d\n", peek(&stack));
    
    display(&stack);
    
    return 0;
}

Applications of stacks

1. Function Call Management (Call Stack)

• Description: The call stack manages function calls and local variables during program execution.
• How It Works:
  • When a function is called, a stack frame (containing return address and local variables) is pushed onto the call stack.
  • When the function returns, the stack frame is popped, and control returns to the calling function.
• Benefits: Helps in managing function calls and recursion, and tracks return addresses.

2. Expression Evaluation

• Description: Stacks are used to evaluate expressions, especially in infix, prefix, and postfix notations.
• Infix Expression: Operators are between operands (e.g., A + B).
• Prefix Expression: Operators precede operands (e.g., + A B).
• Postfix Expression: Operators follow operands (e.g., A B +).
• How It Works:
  • Postfix Evaluation: Uses a stack to store operands until an operator is encountered, which then performs operations on the top operands.
  • Conversion: Infix expressions are converted to postfix using a stack to handle operator precedence and parentheses.

3. Undo Mechanism in Software

• Description: Many software applications (like text editors) use stacks to implement the undo feature.
• How It Works:
  • Each action (like typing or deleting text) is pushed onto a stack.
  • When undo is requested, the most recent action is popped from the stack and reversed.

4. Backtracking Algorithms

• Description: Stacks are used in algorithms that involve exploring multiple possibilities and then reverting changes (backtracking).
• Examples: Maze solving, puzzle solving (like Sudoku or the Eight Queens problem).
• How It Works:
  • As the algorithm explores paths or solutions, it pushes the current state onto a stack.
  • If a path doesn’t work, it pops the stack to revert to a previous state and tries another path.

5. Depth-First Search (DFS) in Graphs

• Description: DFS is a graph traversal algorithm that uses a stack to explore vertices.
• How It Works:
  • Starts at a vertex, pushes it onto the stack, and explores adjacent vertices.
  • Continues until all vertices are explored or the stack is empty.

6. Syntax Parsing

• Description: Stacks are used in syntax parsing to validate and process code in compilers and interpreters.
• How It Works:
  • Parsing: Stack-based parsers (like LL parsers) use stacks to handle nested structures like parentheses and brackets.
  • Balancing Symbols: Used to ensure that symbols (like braces in programming languages) are properly nested and balanced.

7. String Reversal

• Description: Stacks can reverse a string by pushing characters onto the stack and then popping them off in reverse order.
• How It Works:
  • Push each character of the string onto the stack.
  • Pop characters off the stack to get the reversed string.

8. Arithmetic Expression Conversion

• Description: Stacks are used to convert arithmetic expressions from infix to prefix or postfix notation.
• How It Works:
  • Infix to Postfix Conversion: Uses a stack to manage operators and parentheses according to their precedence and associativity rules.

9. Memory Management

• Description: In some memory management schemes, stacks are used to manage memory allocation and deallocation.
• How It Works:
  • Function Call Stack: Keeps track of function calls, local variables, and return addresses.
  • Dynamic Memory Allocation: Some memory allocators use stacks to manage memory blocks and allocation requests.

10. Algorithms and Data Structures

• Description: Several algorithms and data structures rely on stacks.
• Examples:
  • Infix to Postfix Conversion: As mentioned before, converting expressions using stacks.
  • Tower of Hanoi: Solving the Tower of Hanoi problem can be done using a stack-based approach.

Summary

Stacks are fundamental in various domains of computer science, from managing function calls and implementing undo functionality to parsing expressions and exploring algorithms. Their LIFO (Last In, First Out) nature makes them well-suited for scenarios where reversing operations and tracking state changes are essential.

Convert Infix Expression to postfix Expression

The idea is to use the stack data structure to convert an infix expression to a postfix expression. The stack is used to reverse the order of operators in postfix expression. The stack is also used to hold operators since an operator can’t be added to a postfix expression until both of its operands are processed.

Key Concepts

• Infix Expression: Operators are placed between operands (e.g., A + B).
• Postfix Expression: Operators follow their operands (e.g., A B +).

Rules for Conversion

1. Operands (e.g., variables and numbers) are added directly to the output.
2. Operators (e.g., +, -, *, /) are pushed onto the stack, but only after popping operators with higher or equal precedence from the stack.
3. Parentheses are used to override the precedence:
   • Left Parenthesis ((): Push onto the stack.
   • Right Parenthesis ()): Pop from the stack to the output until a left parenthesis is encountered.

Precedence and Associativity

• Precedence: Determines the order of operations. For example, * and / have higher precedence than + and -.
• Associativity: Determines the order of operations for operators with the same precedence. Most operators are left-associative (evaluated left to right), but some (like exponentiation) are right-associative.

Implementation Steps

1. Initialize: Create an empty stack for operators and an empty output list.
2. Read the infix expression: Process each token (operand, operator, or parenthesis).
3. Handle each token:
   • Operand: Add it to the output list.
   • Operator: Pop operators from the stack to the output list while the top of the stack has operators of higher or equal precedence. Push the current operator onto the stack.
   • Left Parenthesis: Push it onto the stack.
   • Right Parenthesis: Pop operators from the stack to the output list until a left parenthesis is encountered.
4. After reading the infix expression: Pop any remaining operators from the stack to the output list.

Algorithm Steps

1. Initialize:
   • An empty stack for operators.
   • An empty list for the output.
2. Read Tokens:
   • If Token is an Operand: Append it to the output list.
   • If Token is a Left Parenthesis (: Push it onto the stack.
   • If Token is a Right Parenthesis ): Pop from the stack to the output list until a left parenthesis is encountered. Discard the parentheses.
   • If Token is an Operator:
     • Pop from the stack to the output list any operators that have higher or equal precedence than the current operator.
     • Push the current operator onto the stack.
3. End of Expression:
   • Pop all operators from the stack to the output list.

To illustrate how the stack performs during the conversion of the infix expression A * (B * C + D * E) + F to postfix notation using boxes, we’ll walk through each token of the expression and show the state of the stack and output at each step.

Conversion Steps with Stack and Output Visualization

Infix Expression: A * (B * C + D * E) + F

Goal: Convert to Postfix Expression

1. Initial State:

• Stack: (empty)
• Output: (empty)

2. Process Each Token:

Token: A

• Action: Operand. Add directly to the output.
• Output: A
• Stack: (empty)

Token: *

• Action: Operator. Push onto the stack.
• Output: A
• Stack:markdownCopy code*

Token: (

• Action: Left parenthesis. Push onto the stack.
• Output: A
• Stack:markdownCopy code* (

Token: B

• Action: Operand. Add directly to the output.
• Output: A B
• Stack:markdownCopy code* (

Token: *

• Action: Operator. Push onto the stack (inside parentheses).
• Output: A B
• Stack:markdownCopy code* ( *

Token: C

• Action: Operand. Add directly to the output.
• Output: A B C
• Stack:markdownCopy code* ( *

Token: +

• Action: Operator. Pop * from the stack to the output (since + has lower precedence). Push + onto the stack.
• Output: A B C *
• Stack:markdownCopy code* ( +

Token: D

• Action: Operand. Add directly to the output.
• Output: A B C * D
• Stack:markdownCopy code* ( +

Token: *

• Action: Operator. Push onto the stack (since * has higher precedence than +).
• Output: A B C * D
• Stack:markdownCopy code* ( + *

Token: E

• Action: Operand. Add directly to the output.
• Output: A B C * D E
• Stack:markdownCopy code* ( + *

Token: )

• Action: Right parenthesis. Pop all operators from the stack until a left parenthesis is encountered.
• Output: A B C * D E * +
• Stack:markdownCopy code*

Token: +

• Action: Operator. Pop * from the stack to the output (since + has lower precedence). Push + onto the stack.
• Output: A B C * D E * + *
• Stack:diffCopy code+

Token: F

• Action: Operand. Add directly to the output.
• Output: A B C * D E * + F
• Stack:diffCopy code+

3. Finish Processing:

• Action: Pop all remaining operators from the stack and add them to the output.
• Output: A B C * D E * + F +
• Stack: (empty)

Summary of Stack and Output States

Final Postfix Expression

Postfix Expression: A B C * D E * + F +

This step-by-step box-based visualization helps clarify how each token affects the stack and the output list, leading to the correct postfix notation.

Program for Conversion of Infix Expression to postfix Expression

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>  // For isdigit function

#define MAX 100

// Stack data structure for operators
typedef struct {
    int top;
    char items[MAX];
} Stack;

// Function prototypes
void initStack(Stack* s);
int isEmpty(Stack* s);
void push(Stack* s, char c);
char pop(Stack* s);
char peek(Stack* s);
int precedence(char op);
int isOperator(char c);
void infixToPostfix(const char* infix, char* postfix);

void initStack(Stack* s) {
    s->top = -1;
}

int isEmpty(Stack* s) {
    return s->top == -1;
}

void push(Stack* s, char c) {
    if (s->top < (MAX - 1)) {
        s->items[++s->top] = c;
    }
}

char pop(Stack* s) {
    if (!isEmpty(s)) {
        return s->items[s->top--];
    }
    return '\0'; // Return null character if stack is empty
}

char peek(Stack* s) {
    if (!isEmpty(s)) {
        return s->items[s->top];
    }
    return '\0'; // Return null character if stack is empty
}

int precedence(char op) {
    switch (op) {
        case '+':
        case '-': return 1;
        case '*':
        case '/': return 2;
        case '^': return 3;
        default: return 0;
    }
}

int isOperator(char c) {
    return c == '+' || c == '-' || c == '*' || c == '/' || c == '^';
}

void infixToPostfix(const char* infix, char* postfix) {
    Stack s;
    initStack(&s);
    int i = 0, k = 0;

    while (infix[i] != '\0') {
        if (isspace(infix[i])) {
            i++;
            continue; // Skip spaces
        }

        if (isdigit(infix[i])) {
            // Process number (operand)
            while (isdigit(infix[i])) {
                postfix[k++] = infix[i++];
            }
            postfix[k++] = ' '; // Separate operands by space
        } else if (infix[i] == '(') {
            push(&s, infix[i++]);
        } else if (infix[i] == ')') {
            while (!isEmpty(&s) && peek(&s) != '(') {
                postfix[k++] = pop(&s);
                postfix[k++] = ' ';
            }
            pop(&s); // Pop the '(' from the stack
            i++;
        } else if (isOperator(infix[i])) {
            while (!isEmpty(&s) && precedence(peek(&s)) >= precedence(infix[i])) {
                postfix[k++] = pop(&s);
                postfix[k++] = ' ';
            }
            push(&s, infix[i++]);
        } else {
            printf("Invalid character %c\n", infix[i]);
            exit(EXIT_FAILURE);
        }
    }

    // Pop remaining operators from the stack
    while (!isEmpty(&s)) {
        postfix[k++] = pop(&s);
        postfix[k++] = ' ';
    }
    postfix[k - 1] = '\0'; // Remove the last space and null-terminate the string
}

int main() {
    char infix[MAX] = "A * ( B * C + D * E ) + F";
    char postfix[MAX];

    infixToPostfix(infix, postfix);
    printf("Postfix Expression: %s\n", postfix);

    return 0;
}

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>  // For isdigit function

#define MAX 100

typedef struct {
    int top;
    char items[MAX];
} Stack;

void initStack(Stack* s);
int isEmpty(Stack* s);
void push(Stack* s, char c);
char pop(Stack* s);
char peek(Stack* s);

int precedence(char op);
int isOperator(char c);

void infixToPostfix(const char* infix, char* postfix);

Stack s;
initStack(&s);
int i = 0, k = 0;

while (infix[i] != '\0') {
    if (isspace(infix[i])) {
        i++;
        continue; // Skip spaces
    }

    if (isdigit(infix[i])) {
        while (isdigit(infix[i])) {
            postfix[k++] = infix[i++];
        }
        postfix[k++] = ' '; // Separate operands by space
    } else if (infix[i] == '(') {
        push(&s, infix[i++]);
    } else if (infix[i] == ')') {
        while (!isEmpty(&s) && peek(&s) != '(') {
            postfix[k++] = pop(&s);
            postfix[k++] = ' ';
        }
        pop(&s); // Pop the '(' from the stack
        i++;
    } else if (isOperator(infix[i])) {
        while (!isEmpty(&s) && precedence(peek(&s)) >= precedence(infix[i])) {
            postfix[k++] = pop(&s);
            postfix[k++] = ' ';
        }
        push(&s, infix[i++]);
    } else {
        printf("Invalid character %c\n", infix[i]);
        exit(EXIT_FAILURE);
    }
}

while (!isEmpty(&s)) {
    postfix[k++] = pop(&s);
    postfix[k++] = ' ';
}
postfix[k - 1] = '\0'; // Remove the last space and null-terminate the string

int main() {
    char infix[MAX] = "A * ( B * C + D * E ) + F";
    char postfix[MAX];

    infixToPostfix(infix, postfix);
    printf("Postfix Expression: %s\n", postfix);

    return 0;
}

Evaluate a Postfix Expression

Evaluating a postfix expression (also known as Reverse Polish Notation) involves using a stack to process operands and operators. The postfix expression eliminates the need for parentheses to denote operation precedence, simplifying evaluation. Here’s a detailed explanation and a C program example for evaluating postfix expressions.

Steps to Evaluate a Postfix Expression

1. Initialize an empty stack.
2. Scan the postfix expression from left to right.
3. Process Each Token:
   • If the token is an operand (number), push it onto the stack.
   • If the token is an operator (e.g., +, -, *, /):
     • Pop the required number of operands from the stack.
     • Perform the operation using these operands.
     • Push the result back onto the stack.
4. After scanning the expression, the final result will be the only element left on the stack.

Algorithm to Evaluate Postfix Expression

1. Initialize an empty stack.
2. Scan the postfix expression from left to right.
3. For each token in the postfix expression:
   • If the token is an operand (number):
     • Push it onto the stack.
   • If the token is an operator (e.g., +, -, *, /):
     • Pop the required number of operands from the stack.
     • Perform the operation with these operands.
     • Push the result back onto the stack.
4. After processing all tokens, the stack should contain only one element, which is the result of the postfix expression.
5. Return the result from the stack.

Postfix Expression: 3 4 * 5 6 * +

Let’s represent the state of the stack and the process for each step:

Initial State

• Stack: (empty)
• Expression: 3 4 * 5 6 * +

Step-by-Step Process

1. Token: 3

• Action: Push 3 onto the stack.
• Stack:| 3 |
• Expression Remaining: 4 * 5 6 * +

2. Token: 4

• Action: Push 4 onto the stack.
• Stack:| 4 | | 3 |
• Expression Remaining: * 5 6 * +

3. Token: *

• Action: Pop 4 and 3, compute 3 * 4 = 12, push 12.
• Stack:| 12 |
• Expression Remaining: 5 6 * +

4. Token: 5

• Action: Push 5 onto the stack.
• Stack:| 5 | | 12 |
• Expression Remaining: 6 * +

5. Token: 6

• Action: Push 6 onto the stack.
• Stack:| 6 | | 5 | | 12 |
• Expression Remaining: * +

6. Token: *

• Action: Pop 6 and 5, compute 5 * 6 = 30, push 30.
• Stack:| 30 | | 12 |
• Expression Remaining: +

7. Token: +

• Action: Pop 30 and 12, compute 12 + 30 = 42, push 42.
• Stack:| 42 |
• Expression Remaining: (empty)

Final State:

• Stack: 42
• Result: 42

Summary of Algorithm with Boxes

1. Initialize: Start with an empty stack.
2. Processing Tokens:
   • For operands, push onto the stack.
   • For operators, pop the required number of operands from the stack, perform the operation, and push the result.
3. Final Result: The last remaining value on the stack is the result.

Example program to Evaluate Postfix Expression

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>  // For isdigit function

#define MAX 100

// Stack data structure for operands
typedef struct {
    int top;
    int items[MAX];
} Stack;

// Function prototypes
void initStack(Stack* s);
int isEmpty(Stack* s);
void push(Stack* s, int value);
int pop(Stack* s);
int peek(Stack* s);
int evaluatePostfix(const char* postfix);

void initStack(Stack* s) {
    s->top = -1;  // Initialize the stack as empty
}

int isEmpty(Stack* s) {
    return s->top == -1;  // Check if the stack is empty
}

void push(Stack* s, int value) {
    if (s->top < (MAX - 1)) {
        s->items[++s->top] = value;  // Push value onto the stack
    }
}

int pop(Stack* s) {
    if (!isEmpty(s)) {
        return s->items[s->top--];  // Pop and return the top value
    }
    return 0;  // Return 0 if stack is empty (should handle this better in production code)
}

int peek(Stack* s) {
    if (!isEmpty(s)) {
        return s->items[s->top];  // Return the top value without removing it
    }
    return 0;  // Return 0 if stack is empty
}

int evaluatePostfix(const char* postfix) {
    Stack s;
    initStack(&s);
    int i = 0;

    while (postfix[i] != '\0') {
        // Skip spaces
        if (isspace(postfix[i])) {
            i++;
            continue;
        }

        // Process number (operand)
        if (isdigit(postfix[i])) {
            int num = 0;
            while (isdigit(postfix[i])) {
                num = num * 10 + (postfix[i] - '0');  // Build the number
                i++;
            }
            push(&s, num);  // Push the number onto the stack
        } 
        // Process operator
        else if (postfix[i] == '+') {
            int op2 = pop(&s);  // Pop two operands
            int op1 = pop(&s);
            push(&s, op1 + op2);  // Perform operation and push result
            i++;
        } 
        else if (postfix[i] == '-') {
            int op2 = pop(&s);
            int op1 = pop(&s);
            push(&s, op1 - op2);
            i++;
        } 
        else if (postfix[i] == '*') {
            int op2 = pop(&s);
            int op1 = pop(&s);
            push(&s, op1 * op2);
            i++;
        } 
        else if (postfix[i] == '/') {
            int op2 = pop(&s);
            int op1 = pop(&s);
            push(&s, op1 / op2);
            i++;
        } 
        else {
            printf("Invalid character %c\n", postfix[i]);
            exit(EXIT_FAILURE);
        }
    }

    return pop(&s);  // The final result is the only element left in the stack
}

int main() {
    char postfix[MAX] = "3 4 * 5 6 * +";  // Postfix expression
    int result = evaluatePostfix(postfix);  // Evaluate the expression
    printf("Result of Postfix Expression: %d\n", result);  // Print the result
    return 0;
}

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>  // For isdigit function

#define MAX 100

typedef struct {
    int top;
    int items[MAX];
} Stack;

void initStack(Stack* s);
int isEmpty(Stack* s);
void push(Stack* s, int value);
int pop(Stack* s);
int peek(Stack* s);

int evaluatePostfix(const char* postfix);

int main() {
    char postfix[MAX] = "3 4 * 5 6 * +";
    int result = evaluatePostfix(postfix);
    printf("Result of Postfix Expression: %d\n", result);
    return 0;
}