Pop Operation on Stack in DSA Java - 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 when the stack grows bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public class Stack {
    private int[] data;
    private int top = -1;

    public int pop() {
        if (top == -1) throw new RuntimeException("Stack is empty");
        return data[top--];
    }
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Accessing and decreasing the top index by one.
• How many times: Exactly once per pop call.

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	1
100	1
1000	1

Pattern observation: The time stays the same even if the stack grows larger.

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: "Popping takes longer if the stack has more items because it has to check all of them."

[OK] Correct: The pop operation only looks at the top item, so it does not check all items. It always takes the same short time.

Interview Connect

Knowing that pop is a quick, constant-time operation helps you explain why stacks are efficient for certain tasks like undo actions or expression evaluation.

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

easy

A. Removes and returns the top element of the stack

B. Adds an element to the bottom of the stack

C. Returns the bottom element without removing it

D. Checks if the stack is empty

Solution

1. Step 1: Understand the stack structure

   A stack follows Last In First Out (LIFO) principle, so operations affect the top element.

2. Step 2: Define pop operation

   The pop() removes and returns the top element from the stack.

3. Final Answer:

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

4. Quick Check:

   pop() removes top element [OK]

Hint: Pop always removes the last added item from stack [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 Java?

easy

A. stack.pop();

B. pop(stack);

C. stack.remove();

D. pop();

Solution

1. Step 1: Recall Java Stack method syntax

   In Java, pop() is a method called on the stack object.

2. Step 2: Identify correct method call

   The correct syntax is calling pop() on the stack instance: stack.pop();

3. Final Answer:

   stack.pop(); -> Option A

4. Quick Check:

   Method call on object = stack.pop() [OK]

Hint: Pop is called as a method on the stack object [OK]

Common Mistakes:

• Using pop() without object
• Using remove() instead of pop()
• Passing stack as argument to pop

3. What will be the output after executing the following Java code?

Stack<Integer> stack = new Stack<>();
stack.push(10);
stack.push(20);
stack.push(30);
int popped = stack.pop();
System.out.println(popped);
System.out.println(stack);

medium

A. 30
[20, 10]

B. 10
[20, 30]

C. 20
[10, 30]

D. 30
[10, 20]

Solution

1. Step 1: Trace push operations

   Elements pushed in order: 10, then 20, then 30. Stack top is 30.

2. Step 2: Apply pop operation

   Pop removes top element 30, so popped = 30. Remaining stack is [10, 20].

3. Final Answer:

   30 [10, 20] -> Option D

4. Quick Check:

   pop() returns last pushed = 30 [OK]

Hint: Pop returns last pushed element, stack shrinks from top [OK]

Common Mistakes:

• Confusing order of elements after pop
• Printing popped element incorrectly
• Assuming pop removes bottom element

4. Consider this Java code snippet:

Stack<String> stack = new Stack<>();
stack.push("A");
stack.push("B");
stack.pop();
stack.pop();
stack.pop();

What error will occur when running this code?

medium

A. No error, code runs fine

B. EmptyStackException at third pop

C. NullPointerException at second pop

D. IndexOutOfBoundsException at first pop

Solution

1. Step 1: Count push and pop calls

   Two elements pushed: "A", "B". Then three pops called.

2. Step 2: Identify error on extra pop

   First two pops remove "B" and "A". Third pop tries to remove from empty stack causing EmptyStackException.

3. Final Answer:

   EmptyStackException at third pop -> Option B

4. Quick Check:

   Pop on empty stack throws EmptyStackException [OK]

Hint: Popping empty stack causes EmptyStackException [OK]

Common Mistakes:

• Assuming pop returns null on empty stack
• Confusing exception types
• Ignoring extra pop call

5. You have a stack with elements [5, 10, 15, 20] (5 at bottom, 20 at top). You perform two pop operations, then push 25, then pop once. What is the final stack content?

hard

A. [5, 10, 25]

B. [5, 10, 15]

C. [5, 10]

D. [5, 10, 15, 25]

Solution

1. Step 1: Perform two pop operations

   Initial stack: bottom 5,10,15, top 20. First pop removes 20, second pop removes 15. Stack now: [5, 10].

2. Step 2: Push 25 and pop once

   Push 25: stack becomes [5, 10, 25]. Pop removes 25, stack returns to [5, 10].

3. Step 3: Verify final stack

   Final stack is [5, 10].

4. Final Answer:

   [5, 10] -> Option C

5. Quick Check:

   Pop removes top, push adds top [OK]

Hint: Track pushes and pops stepwise to find final stack [OK]

Common Mistakes:

• Forgetting last pop removes pushed element
• Mixing order of elements
• Assuming push adds at bottom