Pop Operation on Stack in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output after popping from the stack?

Consider a stack implemented as an array with the following operations. What is the printed stack after one pop operation?

DSA C

int stack[5] = {10, 20, 30, 40, 50};
int top = 4; // points to the last element
// Pop operation
if(top >= 0) {
    top--;
}
// Print stack from bottom to top
for(int i = 0; i <= top; i++) {
    printf("%d ", stack[i]);
}

A 10 20 30 40

B 10 20 30 40 50

C 20 30 40 50

D 10 20 30

💡 Hint

Remember, pop removes the top element from the stack.

The top index starts at 4, pointing to 50. After pop, top decreases to 3, so the stack elements from index 0 to 3 are printed: 10 20 30 40.

❓ Predict Output

intermediate

What happens if we pop from an empty stack?

Given the following code, what is the output?

DSA C

int stack[3] = {0};
int top = -1; // empty stack
// Pop operation
if(top >= 0) {
    top--;
} else {
    printf("Stack Underflow\n");
}

A No output

B 0

C Segmentation fault

D Stack Underflow

💡 Hint

Check the condition before popping.

Since top is -1, the stack is empty. The else block prints 'Stack Underflow'.

🔧 Debug

advanced

Identify the error in this pop operation code

What error will this code produce when popping from the stack?

DSA C

int stack[3] = {1, 2, 3};
int top = 0;
// Pop operation
if(top >= 0) {
    top--;
}
printf("Top element: %d\n", stack[top]);

A Undefined behavior due to incorrect pop condition

B Prints 2

C Prints 1 but pop is not done correctly

D Prints 1

💡 Hint

Check the condition for popping carefully.

The condition 'top > 0' prevents popping when top is 0, so pop never happens. The code prints stack[0] which is 1, but the pop operation is incorrect because it should allow popping when top is 0.

❓ Predict Output

advanced

What is the stack content after multiple pops?

Given the stack and operations below, what is the printed stack?

DSA C

int stack[6] = {5, 10, 15, 20, 25, 30};
int top = 5;
// Pop twice
for(int i = 0; i < 2; i++) {
    if(top >= 0) {
        top--;
    }
}
// Print stack
for(int i = 0; i <= top; i++) {
    printf("%d ", stack[i]);
}

A 15 20 25 30

B 5 10 15 20

C 5 10 15 20 25 30

D 5 10 15

💡 Hint

Each pop reduces the top index by one.

Starting top=5, after two pops top=3. Stack elements from 0 to 3 are printed: 5 10 15 20.

🧠 Conceptual

expert

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

Choose the correct time complexity for the pop operation on a stack implemented using an array.

A O(1) - Constant time

B O(n) - Linear time

C O(log n) - Logarithmic time

D O(n^2) - Quadratic time

💡 Hint

Pop just moves the top pointer.

Pop operation only decrements the top index, which is a constant time operation.