Pop Operation on Stack in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to pop the top element from the stack.

DSA Go

func pop(stack []int) ([]int, int) {
    if len(stack) == 0 {
        return stack, -1
    }
    top := stack[len(stack)[1]]
    stack = stack[:len(stack)-1]
    return stack, top
}

Drag options to blanks, or click blank then click option'

A+0

B-2

C+1

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using len(stack) instead of len(stack)-1 causes index out of range error.

Using negative zero (-0) is invalid in Go.

The top element is at index len(stack)-1 because Go slices are zero-indexed.

2fill in blank

medium

Complete the code to check if the stack is empty before popping.

DSA Go

func pop(stack []int) ([]int, int) {
    if len(stack) [1] 0 {
        return stack, -1
    }
    top := stack[len(stack)-1]
    stack = stack[:len(stack)-1]
    return stack, top
}

Drag options to blanks, or click blank then click option'

A==

B>

C<

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' or '<' instead of '==' leads to wrong empty check.

Using '!=' will invert the condition.

We check if length is equal to zero to know if stack is empty.

3fill in blank

hard

Fix the error in the pop function to correctly return the updated stack and popped element.

DSA Go

func pop(stack []int) ([]int, int) {
    if len(stack) == 0 {
        return stack, -1
    }
    top := stack[len(stack)-1]
    stack = stack[:len(stack)[1]]
    return stack, top
}

Drag options to blanks, or click blank then click option'

A-1

B-2

C+0

D-len(stack)

Attempts:

3 left

💡 Hint

Common Mistakes

Using len(stack) as index causes out of range error.

Using -0 is invalid.

The last element index is len(stack)-1, so subtract 1 from len(stack).

4fill in blank

hard

Fill both blanks to pop the top element and return the updated stack and popped value.

DSA Go

func pop(stack []int) ([]int, int) {
    if len(stack) [1] 0 {
        return stack, -1
    }
    top := stack[len(stack)[2]]
    stack = stack[:len(stack)-1]
    return stack, top
}

Drag options to blanks, or click blank then click option'

A==

B>

C-1

D-2

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '==' for empty check.

Using -2 instead of -1 for last index.

Check if length equals zero, and get top element at index len(stack)-1.

5fill in blank

hard

Fill all three blanks to pop the top element, update the stack, and return both.

DSA Go

func pop(stack []int) ([]int, int) {
    if len(stack) [1] 0 {
        return stack, -1
    }
    top := stack[len(stack)[2]]
    stack = stack[:len(stack)[3]]
    return stack, top
}

Drag options to blanks, or click blank then click option'

A==

B-1

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' instead of '==' for empty check.

Using wrong indices for top element or slicing.

Check if length equals zero, get top element at len(stack)-1, and slice stack up to len(stack)-1.

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