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

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.

How does the time needed change as the stack grows bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Remove the top item from the stack
int pop() {
  if (isEmpty()) {
    throw std::out_of_range("Stack is empty");
  }
  int topItem = stackArray[topIndex];
  topIndex--;
  return topItem;
}
    

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: This happens once per pop call, no loops or recursion.

How Execution Grows With Input

Removing the top item always takes the same steps, no matter how big the stack is.

Input Size (n)	Approx. Operations
10	3 steps (check empty, access top, decrease index)
100	3 steps (same as above)
1000	3 steps (same as above)

Pattern observation: The number of steps stays the same no matter how large the stack is.

Final Time Complexity

Time Complexity: O(1)

This means popping an item takes a fixed amount of time regardless of stack size.

Common Mistake

[X] Wrong: "Pop takes longer if the stack is bigger because there are more items."

[OK] Correct: Pop only removes the top item directly, so it does not depend on stack size.

Interview Connect

Knowing that pop is a quick, constant-time operation helps you explain stack efficiency clearly in interviews.

Self-Check

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

Practice

1. What does the pop() operation do in a stack?

easy

A. Adds a new element to the bottom of the stack

B. Removes the top element from the stack

C. Returns the size 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 removed first.

2. Step 2: Define pop operation

   The pop operation removes the top element from the stack, which is the last inserted item.

3. Final Answer:

   Removes the top element from the stack -> Option B

4. Quick Check:

   pop() removes top element [OK]

Hint: Pop always removes the last added item [OK]

Common Mistakes:

• Thinking pop adds elements
• Confusing pop with peek or size
• Assuming pop clears the stack

2. Which of the following is the correct syntax to pop an element from a stack named stk in C++ STL?

easy

A. stk.pop();

B. pop(stk);

C. stk.pop_element();

D. stk.remove();

Solution

1. Step 1: Recall C++ STL stack syntax

   The standard stack class in C++ uses member functions like push() and pop() called on the stack object.

2. Step 2: Identify correct pop syntax

   The correct way to remove the top element is calling pop() as stk.pop();.

3. Final Answer:

   stk.pop(); -> Option A

4. Quick Check:

   Use object.pop() to remove top [OK]

Hint: Use object.pop() to remove top element [OK]

Common Mistakes:

• Using pop() as a free function
• Using incorrect method names like pop_element()
• Trying to remove with remove() which doesn't exist

3. What will be the output of the following C++ code?

std::stack<int> stk;
stk.push(10);
stk.push(20);
stk.pop();
std::cout << stk.top();

medium

A. 0

B. 20

C. 10

D. Compilation error

Solution

1. Step 1: Trace push operations

   First, 10 is pushed, then 20 is pushed on top. Stack top is now 20.

2. Step 2: Apply pop and check top

   pop() removes the top element (20), so top becomes 10. Printing top outputs 10.

3. Final Answer:

   10 -> Option C

4. Quick Check:

   After pop, top is 10 [OK]

Hint: Pop removes last pushed, top shows new top [OK]

Common Mistakes:

• Assuming top remains 20 after pop
• Confusing pop with peek
• Expecting 0 or error output

4. Identify the error in the following code snippet that uses a stack:

std::stack<int> stk;
stk.pop();
stk.push(5);

medium

A. No error, code runs fine

B. Push should be called before pop

C. Stack declaration is incorrect

D. Pop called on empty stack causes runtime error

Solution

1. Step 1: Analyze stack state before pop

   The stack is empty initially. Calling pop() on empty stack is invalid and causes runtime error.

2. Step 2: Check order of operations

   Pop is called before any push, so no element exists to remove, causing error.

3. Final Answer:

   Pop called on empty stack causes runtime error -> Option D

4. Quick Check:

   Pop on empty stack causes error [OK]

Hint: Never pop from empty stack [OK]

Common Mistakes:

• Ignoring empty stack before pop
• Thinking push order fixes error
• Assuming pop on empty is safe

5. Given a stack with elements 1, 2, 3, 4 pushed in that order (1 at bottom, 4 at top), what will be the stack content after performing two pop() operations?

hard

A. 1 -> 2 -> null

B. 3 -> 4 -> null

C. 1 -> 2 -> 3 -> null

D. 4 -> 3 -> 2 -> 1 -> null

Solution

1. Step 1: Understand initial stack order

   Elements pushed in order: 1 (bottom), 2, 3, 4 (top). Stack looks like 1 -> 2 -> 3 -> 4 -> null.

2. Step 2: Apply two pop operations

   First pop removes 4 (top), second pop removes 3 (new top). Remaining stack is 1 -> 2 -> null.

3. Final Answer:

   1 -> 2 -> null -> Option A

4. Quick Check:

   Two pops remove top two elements [OK]

Hint: Pop removes top elements one by one [OK]

Common Mistakes:

• Confusing order of elements after pop
• Listing popped elements instead of remaining
• Reversing stack order incorrectly