Pop Operation on Stack in DSA Go - 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 using the pop operation.

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Pop removes the top element from the stack
func (s *Stack) Pop() (int, bool) {
    if len(s.items) == 0 {
        return 0, false // stack empty
    }
    topIndex := len(s.items) - 1
    topItem := s.items[topIndex]
    s.items = s.items[:topIndex]
    return topItem, true
}
    

This code removes the last element added to the stack and returns it.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Accessing and slicing the last element of the stack's internal slice.
• How many times: This happens once per pop call, no loops or recursion involved.

How Execution Grows With Input

Removing the top item always takes the same steps regardless of stack size.

Input Size (n)	Approx. Operations
10	5 steps
100	5 steps
1000	5 steps

Pattern observation: The number of steps stays constant 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 when the stack is bigger because there are more items to check."

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

Interview Connect

Knowing that pop is a quick, constant-time operation helps you explain why stacks are efficient for last-in-first-out tasks.

Self-Check

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

Practice

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

easy

A. Clears all elements from the stack

B. Adds an element to the bottom of the stack

C. Returns the bottom element without removing it

D. Removes and returns the top element of 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 and returns the top element of the stack, which is the last inserted item.

3. Final Answer:

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

4. Quick Check:

   Pop = remove top element [OK]

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

Common Mistakes:

• Thinking pop removes from bottom
• Confusing pop with peek
• Assuming pop clears the stack

2. Which of the following is the correct Go syntax to pop an element from a stack represented as a slice named stack?

easy

A. element := stack[0]; stack = stack[1:]

B. element := stack[len(stack)-1]; stack = stack[:len(stack)-1]

C. element := stack[len(stack)]; stack = stack[:len(stack)-1]

D. element := stack[len(stack)-1]; stack = stack[1:]

Solution

1. Step 1: Identify top element index in slice

   In Go, the top of the stack is the last element: index len(stack)-1.

2. Step 2: Correctly slice to remove top element

   To pop, get element at len(stack)-1, then slice stack up to len(stack)-1 to remove it.

3. Final Answer:

   element := stack[len(stack)-1]; stack = stack[:len(stack)-1] -> Option B

4. Quick Check:

   Pop syntax = element last, slice off last [OK]

Hint: Pop last element with slice[:len(slice)-1] [OK]

Common Mistakes:

• Using index len(stack) which is out of range
• Removing from front instead of end
• Not updating the stack slice after pop

3. What is the output of the following Go code?

stack := []int{10, 20, 30}
element := stack[len(stack)-1]
stack = stack[:len(stack)-1]
fmt.Println(element)
fmt.Println(stack)

medium

A. 30
[10 20]

B. 20
[10 20 30]

C. 30
[10 20 30]

D. 10
[20 30]

Solution

1. Step 1: Identify popped element

   element = stack[len(stack)-1] = stack[2] = 30

2. Step 2: Update stack after pop

   stack = stack[:2] = [10, 20]

3. Final Answer:

   30 [10 20] -> Option A

4. Quick Check:

   Pop returns 30, stack left [10 20] [OK]

Hint: Pop returns last element, stack shrinks by one [OK]

Common Mistakes:

• Printing old stack without slicing
• Confusing index for popped element
• Not updating stack slice after pop

4. Consider this Go code snippet for popping from a stack:

stack := []int{5, 15, 25}
element := stack[len(stack)]
stack = stack[:len(stack)-1]
fmt.Println(element)

What is the error in this code?

medium

A. Index out of range error due to accessing stack[len(stack)]

B. Stack slice is not updated correctly after pop

C. Variable element is not declared

D. No error, code runs fine

Solution

1. Step 1: Check index used for popping

   Accessing stack[len(stack)] is invalid because slice indices go from 0 to len(stack)-1.

2. Step 2: Identify runtime error

   This causes an index out of range runtime error when trying to access stack[len(stack)].

3. Final Answer:

   Index out of range error due to accessing stack[len(stack)] -> Option A

4. Quick Check:

   Index must be len(stack)-1, not len(stack) [OK]

Hint: Slice indices max at len(slice)-1, not len(slice) [OK]

Common Mistakes:

• Using len(stack) as index
• Ignoring runtime errors
• Assuming slice indexing is inclusive of len

5. You have a stack implemented as a slice in Go: stack := []int{1, 2, 3, 4, 5}. You want to pop two elements and then push 10 and 20. What will be the final stack?

hard

A. [10 20]

B. [1 2 3 4 5 10 20]

C. [1 2 3 10 20]

D. [1 2 10 20]

Solution

1. Step 1: Pop two elements from stack

   Pop removes last two elements: 5 and 4. Stack becomes [1, 2, 3].

2. Step 2: Push 10 and 20 onto stack

   Appending 10 and 20 results in [1, 2, 3, 10, 20].

3. Final Answer:

   [1 2 3 10 20] -> Option C

4. Quick Check:

   Pop twice, then push twice = [1 2 3 10 20] [OK]

Hint: Pop removes last; push adds at end [OK]

Common Mistakes:

• Not popping twice before pushing
• Appending instead of replacing popped elements
• Confusing order of push and pop