Pop Operation on Stack in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the pop operation do in a stack?

It removes the top element from the stack and returns it, reducing the stack size by one.

beginner

What happens if you try to pop from an empty stack?

This is called stack underflow. It means there is no element to remove, and the operation should handle this error.

beginner

In C, which variable usually tracks the top of the stack for pop operation?

An integer variable called top is used to track the index of the top element in the stack array.

beginner

What is the time complexity of the pop operation on a stack?

The pop operation runs in O(1) time because it only removes the top element without traversing the stack.

beginner

Show the basic steps of the pop operation on a stack implemented with an array.

1. Check if the stack is empty (top == -1).
2. If not empty, get the element at stack[top].
3. Decrease top by 1.
4. Return the popped element.

What does the pop operation return when the stack is empty?

A An error or special value indicating underflow

B The bottom element of the stack

C The last pushed element without removing it

D The size of the stack

Pop on an empty stack causes underflow, so it returns an error or special value.

Which variable is commonly used to track the top element in a stack implemented with an array?

A tail

B head

C top

D count

The variable 'top' tracks the index of the top element in the stack.

What is the time complexity of the pop operation on a stack?

A O(n^2)

B O(log n)

C O(n)

D O(1)

Pop removes the top element directly, so it runs in constant time O(1).

After popping an element, what happens to the 'top' variable?

A It decreases by 1

B It stays the same

C It increases by 1

D It resets to zero

The 'top' decreases by 1 to point to the new top element after pop.

Which of the following is NOT a step in the pop operation?

A Check if stack is empty

B Increase the top index

C Return the top element

D Decrease the top index

Pop decreases the top index; increasing it is incorrect.

Explain the pop operation on a stack implemented using an array.

Think about what happens when you remove the last added item.

Describe what stack underflow means and how it relates to the pop operation.

Consider what happens if you pop too many times.

Practice

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