Pop Operation on Stack in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the pop operation do in a stack?

The pop operation removes the top element from the stack and returns it. If the stack is empty, it usually returns an error or a special value.

beginner

In a stack, which end do we perform the pop operation on?

We perform the pop operation on the top end of the stack, following the Last In First Out (LIFO) principle.

intermediate

What happens if you try to pop from an empty stack?

Trying to pop from an empty stack causes an underflow condition. It means there is no element to remove, and the operation should handle this safely, often by returning an error.

beginner

Show the state of the stack after popping once from this stack: 5 -> 10 -> 15 -> null

After popping once, the top element 15 is removed. The stack becomes: 5 -> 10 -> null

intermediate

Why is the pop operation important in stack-based algorithms?

Pop is important because it allows us to retrieve and remove the most recently added item, enabling backtracking, undo operations, and managing function calls in programming.

What does the pop operation return when performed on a non-empty stack?

A The top element of the stack

B The bottom element of the stack

C The size of the stack

D Nothing

Pop removes and returns the top element of the stack.

What is the state of the stack after popping from: 3 -> 7 -> 9 -> null?

A 7 -> 9 -> null

B Empty stack

C 3 -> 9 -> null

D 3 -> 7 -> null

Pop removes the top element 9, leaving 3 -> 7 -> null.

What is the term for trying to pop from an empty stack?

A Underflow

B Stack error

C Stack full

D Overflow

Underflow means popping from an empty stack.

Which principle does the pop operation follow in a stack?

A First In First Out (FIFO)

B Last In First Out (LIFO)

C Random Access

D Priority Based

Stacks follow LIFO, so pop removes the last added element.

If a stack has elements 1 -> 2 -> 3 -> null, what will be the top element after one pop?

A 3

B 1

C 2

D null

Pop removes 3, so 2 becomes the new top.

Explain the pop operation on a stack and what happens when the stack is empty.

Think about what happens when you try to take something from an empty box.

Describe how the stack changes visually after performing a pop operation on a stack with multiple elements.

Imagine a stack of plates and taking the top plate off.

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