Pop Operation on Stack in DSA Go - Execution Trace

Concept Flow - Pop Operation on Stack

Check if stack is empty?

No↓

Remove top element

↓

Update top pointer

↓

Return removed element

↓

Done

Yes↓

Error: Stack Underflow

The pop operation first checks if the stack is empty. If not, it removes the top element, updates the top pointer, and returns the removed element. If empty, it signals an underflow error.

Execution Sample

DSA Go

stack := [10, 20, 30]
top := 2
pop()
// removes 30, top becomes 1

This code removes the top element (30) from the stack and updates the top pointer.

Execution Table

Step	Operation	Nodes in Stack	Top Pointer	Visual State
1	Check if stack is empty	[10, 20, 30]	2	10 -> 20 -> 30 (top)
2	Remove top element (30)	[10, 20, 30]	2	10 -> 20 -> 30 (removed)
3	Update top pointer to 1	[10, 20, 30]	1	10 -> 20 (top)
4	Return removed element 30	[10, 20, 30]	1	10 -> 20 (top)
5	Done	[10, 20, 30]	1	10 -> 20 (top)

💡 Top pointer updated to 1, effective stack size reduced by one, pop operation complete

Variable Tracker

Variable	Start	After Step 2	After Step 3	Final
stack	[10, 20, 30]	[10, 20, 30]	[10, 20, 30]	[10, 20, 30]
top	2	2	1	1
popped_element	nil	30	30	30

Key Moments - 3 Insights

Why do we check if the stack is empty before popping?

Because popping from an empty stack causes an error (underflow). The execution_table step 1 shows this check to prevent removing from an empty stack.

Does the popped element remain in the stack after pop?

No, the popped element is removed logically. Step 2 shows the element 30 is removed, and the visual state updates to exclude it.

Why do we update the top pointer after removing the element?

The top pointer must point to the new top element after pop. Step 3 updates top from 2 to 1 to reflect the new top.

Visual Quiz - 3 Questions

Look at the execution_table, what is the top pointer value after step 3?

A 1

B 2

C 0

D 3

💡 Hint

Refer to the 'Top Pointer' column in execution_table row for step 3

At which step is the element 30 removed from the stack?

A Step 3

B Step 1

C Step 2

D Step 4

💡 Hint

Check the 'Operation' and 'Visual State' columns in execution_table for when 30 disappears

If the stack was empty at the start, what would happen at step 1?

A Remove top element

B Error: Stack Underflow

C Update top pointer

D Return popped element

💡 Hint

Look at concept_flow and execution_table step 1 for empty stack check

Concept Snapshot

Pop Operation on Stack:
- Check if stack is empty
- Remove top element if not empty
- Update top pointer to new top
- Return removed element
- Error if stack empty (underflow)

Full Transcript

The pop operation on a stack first checks if the stack is empty to avoid errors. If not empty, it removes the top element, updates the top pointer to the next element below, and returns the removed element. If the stack is empty, it signals an underflow error. The execution table shows each step with the stack's visual state and pointer changes. The variable tracker follows the stack array, top pointer, and popped element values through the operation.

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