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DSA Cprogramming~5 mins

Pop Operation on Stack in DSA C - Time & Space Complexity

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Time Complexity: Pop Operation on Stack
O(1)
Understanding Time Complexity

We want to understand how long it takes to remove an item from a stack.

Specifically, how the time changes as the stack grows.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.


// Remove top item from stack if not empty
int pop(Stack* s) {
    if (s->top == -1) {
        return -1; // stack empty
    }
    int item = s->arr[s->top];
    s->top--;
    return item;
}
    

This code removes the top element from the stack and returns it.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

  • Primary operation: Accessing and removing the top element.
  • How many times: Exactly once per pop call.
How Execution Grows With Input

Each pop removes one item with a fixed number of steps.

Input Size (n)Approx. Operations
101
1001
10001

Pattern observation: The time stays the same no matter how big the stack is.

Final Time Complexity

Time Complexity: O(1)

This means popping an item takes the same short time no matter how many items are in the stack.

Common Mistake

[X] Wrong: "Pop takes longer if the stack is bigger because it has to look through all items."

[OK] Correct: Pop only removes the top item directly without checking others, so size does not affect time.

Interview Connect

Knowing that pop is very fast helps you understand stack efficiency and shows you can analyze simple but important operations clearly.

Self-Check

"What if the stack was implemented using a linked list instead of an array? How would the time complexity of pop change?"

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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10 questions~5 mineasy hard