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DSA Cprogramming

Pop Operation on Stack in DSA C

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Mental Model
A stack is like a pile of plates where you only remove the top plate. Pop means taking off the top item.
Analogy: Imagine a stack of books on a table. You can only take the top book off first, not the ones below it.
Top
 ↑
[3] -> [2] -> [1] -> null
Bottom
Dry Run Walkthrough
Input: stack: 3 -> 2 -> 1 -> null, pop operation
Goal: Remove the top element (3) from the stack and update the top pointer
Step 1: Check if stack is empty
Top -> [3] -> [2] -> [1] -> null
Why: We must ensure there is something to pop
Step 2: Save top element (3) to return later
Top -> [3] -> [2] -> [1] -> null
Why: We need to know which element is being removed
Step 3: Move top pointer to next element (2)
Top -> [2] -> [1] -> null
Why: Removing the top means the next element becomes the new top
Step 4: Return saved element (3) as popped value
Top -> [2] -> [1] -> null
Why: Pop operation outputs the removed top element
Result: 
Top -> [2] -> [1] -> null
Popped element: 3
Annotated Code
DSA C
#include <stdio.h>
#include <stdlib.h>

// Node structure for stack
typedef struct Node {
    int data;
    struct Node* next;
} Node;

// Push function to add element to stack
void push(Node** top_ref, int new_data) {
    Node* new_node = (Node*)malloc(sizeof(Node));
    new_node->data = new_data;
    new_node->next = *top_ref;
    *top_ref = new_node;
}

// Pop function to remove top element from stack
int pop(Node** top_ref) {
    if (*top_ref == NULL) {
        printf("Stack Underflow\n");
        return -1; // Indicate empty stack
    }
    Node* temp = *top_ref; // Save top node
    int popped = temp->data; // Save data to return
    *top_ref = temp->next; // Move top to next node
    free(temp); // Free old top node
    return popped; // Return popped data
}

// Print stack from top to bottom
void printStack(Node* top) {
    Node* current = top;
    while (current != NULL) {
        printf("%d -> ", current->data);
        current = current->next;
    }
    printf("null\n");
}

int main() {
    Node* stack = NULL;
    push(&stack, 1);
    push(&stack, 2);
    push(&stack, 3);

    printf("Stack before pop: ");
    printStack(stack);

    int popped = pop(&stack);
    printf("Popped element: %d\n", popped);

    printf("Stack after pop: ");
    printStack(stack);

    return 0;
}
if (*top_ref == NULL) {
Check if stack is empty to avoid popping from empty stack
Node* temp = *top_ref; // Save top node
Save current top node to remove it safely
int popped = temp->data; // Save data to return
Store the data of the top node to return after removal
*top_ref = temp->next; // Move top to next node
Update top pointer to next node, effectively removing top
free(temp); // Free old top node
Free memory of removed node to avoid leaks
return popped; // Return popped data
Return the value popped from the stack
OutputSuccess
Stack before pop: 3 -> 2 -> 1 -> null Popped element: 3 Stack after pop: 2 -> 1 -> null
Complexity Analysis
Time: O(1) because pop removes only the top element without traversal
Space: O(1) as no extra space is needed besides temporary variables
vs Alternative: Compared to array-based stack pop which is also O(1), linked list pop avoids resizing issues
Edge Cases
Empty stack
Prints 'Stack Underflow' and returns -1 to indicate no element to pop
DSA C
if (*top_ref == NULL) {
Single element stack
Pops the only element and sets top to NULL, resulting in empty stack
DSA C
*top_ref = temp->next;
When to Use This Pattern
When you see a problem requiring last-in-first-out removal, reach for the pop operation on stack because it removes the top element efficiently.
Common Mistakes
Mistake: Not checking if the stack is empty before popping
Fix: Add a condition to check if top is NULL before removing to avoid errors
Mistake: Not updating the top pointer after popping
Fix: Set top to next node after removing the current top to maintain stack integrity
Summary
Removes the top element from the stack and returns it.
Use when you need to retrieve and remove the most recently added item.
The key is to update the top pointer to the next element after removal.

Practice

(1/5)
1. What does the pop operation do in a stack?
easy
A. Returns the bottom element without removing it
B. Adds a new element to the bottom of the stack
C. Removes and returns the top element of the stack
D. Clears all elements from the stack

Solution

  1. Step 1: Understand stack behavior

    A stack follows Last In First Out (LIFO) principle, so the last added element is at the top.

  2. Step 2: Define pop operation

    The pop operation removes and returns this top element, reducing the stack size by one.

  3. Final Answer:

    Removes and returns the top element of the stack -> Option C

  4. Quick Check:

    Pop = remove top element [OK]
Hint: Pop always removes the last added item [OK]
Common Mistakes:
  • Thinking pop adds elements
  • Confusing pop with peek
  • Assuming pop removes bottom element
2. Which of the following is the correct syntax to pop an element from a stack named stack in C?
easy
A. pop(&stack);
B. stack.pop();
C. pop(stack);
D. stack.pop;

Solution

  1. Step 1: Understand C function call for pop

    In C, stack is usually a struct or array, so pop function requires address to modify it.

  2. Step 2: Identify correct syntax

    Passing &stack (address) to pop function is correct to modify stack state.

  3. Final Answer:

    pop(&stack); -> Option A

  4. Quick Check:

    Use & to pass stack address in C [OK]
Hint: Use & to pass stack pointer in C functions [OK]
Common Mistakes:
  • Using dot notation like in C++ or Java
  • Calling pop without & causing no stack change
  • Missing parentheses in function call
3. What will be the output after executing the following code?
typedef struct {
  int arr[3];
  int top;
} Stack;

void pop(Stack *s) {
  if (s->top == -1) return;
  s->top--;
}

int main() {
  Stack s = {{10, 20, 30}, 2};
  pop(&s);
  printf("%d", s.arr[s.top]);
  return 0;
}
medium
A. 20
B. 30
C. 10
D. Runtime error

Solution

  1. Step 1: Initial stack state

    Stack has elements [10, 20, 30] with top = 2 (pointing to 30).

  2. Step 2: Perform pop operation

    Pop decreases top from 2 to 1, so top now points to 20.

  3. Step 3: Print element at new top

    Printing s.arr[s.top] prints s.arr[1] which is 20.

  4. Final Answer:

    20 -> Option A

  5. Quick Check:

    Pop reduces top, new top element = 20 [OK]
Hint: Pop moves top down by one, print new top [OK]
Common Mistakes:
  • Printing old top element after pop
  • Confusing array index with top value
  • Ignoring top decrement in pop
4. Identify the error in the following pop function for a stack and select the correct fix:
void pop(Stack *s) {
  if (s->top == 0) {
    printf("Stack Underflow\n");
    return;
  }
  s->top--;
}
medium
A. Add return value to pop function
B. Change condition to s->top == -1 to check empty stack
C. Change s->top-- to s->top++
D. Remove the if condition entirely

Solution

  1. Step 1: Understand empty stack condition

    Stack is empty when top is -1, not 0, because top starts at -1.

  2. Step 2: Fix condition to detect underflow

    Change condition to s->top == -1 to correctly detect empty stack before popping.

  3. Final Answer:

    Change condition to s->top == -1 to check empty stack -> Option B

  4. Quick Check:

    Empty stack top = -1, check before pop [OK]
Hint: Empty stack top is -1, not 0 [OK]
Common Mistakes:
  • Checking top == 0 instead of -1
  • Incrementing top on pop
  • Ignoring underflow check
5. Given a stack implemented with an array of size 5 and top initially -1, after pushing 3 elements and then popping 4 times, what will be the value of top?
hard
A. 0
B. 1
C. -2
D. -1

Solution

  1. Step 1: Calculate top after pushing 3 elements

    Starting top = -1, pushing 3 times increments top to 2 (0,1,2).

  2. Step 2: Calculate top after popping 4 times

    Each pop decrements top by 1: 2 -> 1 -> 0 -> -1. The 4th pop tries to pop empty stack, top stays at -1.

  3. Final Answer:

    -1 -> Option D

  4. Quick Check:

    Pop beyond empty stack keeps top at -1 [OK]
Hint: Top never goes below -1 after pops [OK]
Common Mistakes:
  • Assuming top can go below -1
  • Counting pushes and pops incorrectly
  • Ignoring underflow protection
Recommended

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