Dynamic Stack Using Resizable Array in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to push an element onto the stack.

DSA Go

func (s *Stack) Push(value int) {
    if s.size == len(s.data) {
        s.resize(s.size * 2)
    }
    s.data[s.size] = [1]
    s.size++
}

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

Alen

Bsize

Cdata

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'size' instead of 'value' to assign the element.

Using 'data' which is the slice, not the element.

We push the given value onto the stack at the current size index.

2fill in blank

medium

Complete the code to pop an element from the stack.

DSA Go

func (s *Stack) Pop() (int, bool) {
    if s.size == 0 {
        return 0, false
    }
    s.size--
    value := s.data[[1]]
    s.data[s.size] = 0
    if s.size > 0 && s.size == len(s.data)/4 {
        s.resize(len(s.data) / 2)
    }
    return value, true
}

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

A0

Bs.size - 1

Cs.size

Dlen(s.data)

Attempts:

3 left

💡 Hint

Common Mistakes

Using 's.size - 1' after decrement, causing off-by-one error.

Using 0 or len(s.data) which are incorrect indices.

We access the element at index s.size because size was decremented before.

3fill in blank

hard

Fix the error in the resize method to correctly copy elements to the new array.

DSA Go

func (s *Stack) resize(capacity int) {
    newData := make([]int, capacity)
    for i := 0; i < s.size; i++ {
        newData[i] = [1]
    }
    s.data = newData
}

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

As.data[i]

Bcapacity

CnewData[i]

Ds.size

Attempts:

3 left

💡 Hint

Common Mistakes

Copying from newData instead of s.data causes zero values.

Using capacity or s.size as values instead of elements.

We copy elements from the old data slice at index i to the new slice.

4fill in blank

hard

Fill both blanks to complete the Peek method that returns the top element without removing it.

DSA Go

func (s *Stack) Peek() (int, bool) {
    if s.size == 0 {
        return 0, false
    }
    return s.data[[1]], [2]
}

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

As.size - 1

Btrue

Cfalse

Ds.size

Attempts:

3 left

💡 Hint

Common Mistakes

Using s.size as index causes out-of-range error.

Returning false incorrectly when stack has elements.

Peek returns the element at index size-1 and true to indicate success.

5fill in blank

hard

Fill all three blanks to complete the IsEmpty method that checks if the stack is empty.

DSA Go

func (s *Stack) IsEmpty() bool {
    return [1] == [2] || s.size == [3]
}

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

As.size

B0

Clen(s.data)

Dcapacity

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing size to len(s.data) or capacity which is incorrect.

Using wrong variables for emptiness check.

Stack is empty if size equals zero; len(s.data) or capacity are not used here.

Practice

1. What is the main advantage of using a dynamic stack implemented with a resizable array compared to a fixed-size array stack?

easy

A. It can grow or shrink automatically as elements are pushed or popped.

B. It uses less memory than a linked list stack.

C. It always has a fixed maximum size.

D. It does not require resizing when full.

Solution

1. Step 1: Understand fixed-size array stack limitation

   A fixed-size array stack cannot grow beyond its initial capacity, causing overflow if too many elements are pushed.

2. Step 2: Understand dynamic stack behavior

   A dynamic stack with a resizable array automatically increases or decreases its size when needed, avoiding overflow and wasting less memory.

3. Final Answer:

   It can grow or shrink automatically as elements are pushed or popped. -> Option A

4. Quick Check:

   Dynamic resizing = Automatic growth/shrink [OK]

Hint: Dynamic arrays resize automatically on overflow or underflow [OK]

Common Mistakes:

• Thinking dynamic stack has fixed size
• Confusing linked list stack with array stack
• Assuming no resizing is needed

2. Which of the following Go code snippets correctly doubles the size of the stack's underlying array when it is full?

easy

A. stack.data = append(stack.data, make([]int, len(stack.data))...)

B. newData := make([]int, len(stack.data)*2); copy(newData, stack.data); stack.data = newData

C. stack.data = stack.data[:len(stack.data)*2]

D. stack.data = make([]int, len(stack.data)/2)

Solution

1. Step 1: Understand how to resize an array in Go

   Go slices cannot be resized directly; we create a new slice with double capacity and copy old elements.

2. Step 2: Check each option

   newData := make([]int, len(stack.data)*2); copy(newData, stack.data); stack.data = newData creates a new slice double the size, copies old data, and assigns it back, which is correct resizing.

3. Final Answer:

   newData := make([]int, len(stack.data)*2); copy(newData, stack.data); stack.data = newData -> Option B

4. Quick Check:

   Resize by new slice + copy = newData := make([]int, len(stack.data)*2); copy(newData, stack.data); stack.data = newData [OK]

Hint: Resize by new slice + copy old data [OK]

Common Mistakes:

• Using append to resize underlying array incorrectly
• Slicing beyond capacity without copying
• Reducing size instead of increasing

3. Consider the following Go code for a dynamic stack push operation:

func (s *Stack) Push(val int) {
  if s.size == len(s.data) {
    newData := make([]int, len(s.data)*2)
    copy(newData, s.data)
    s.data = newData
  }
  s.data[s.size] = val
  s.size++
}

stack := Stack{data: make([]int, 2), size: 0}
stack.Push(10)
stack.Push(20)
stack.Push(30)
fmt.Println(stack.data[:stack.size])

What is the output?

medium

A. [10 20]

B. [10 20 0]

C. [10 20 30]

D. runtime error: index out of range

Solution

1. Step 1: Trace initial stack state and pushes

   Initial data slice size is 2, size=0. Push 10 at index 0, size=1; Push 20 at index 1, size=2.

2. Step 2: Push 30 triggers resize

   Since size == len(data), create newData with size 4, copy old data, assign newData. Then add 30 at index 2, size=3.

3. Step 3: Print slice up to size

   Print data[:3] which contains [10 20 30].

4. Final Answer:

   [10 20 30] -> Option C

5. Quick Check:

   Resize + push = [10 20 30] [OK]

Hint: Resize before push when full, then add element [OK]

Common Mistakes:

• Forgetting to resize before push
• Printing full data slice including zeros
• Index out of range by pushing without resize

4. The following Go code for popping from a dynamic stack sometimes causes a runtime error. What is the bug?

func (s *Stack) Pop() int {
  s.size--
  val := s.data[s.size]
  if s.size < len(s.data)/4 {
    newData := make([]int, len(s.data)/2)
    copy(newData, s.data[:s.size])
    s.data = newData
  }
  return val
}

medium

A. Decreasing size before checking if stack is empty causes negative index.

B. Resizing array to half size before copying causes data loss.

C. Copying data after resizing causes runtime error.

D. Returning val before resizing causes stale value.

Solution

1. Step 1: Analyze size decrement before access

   Decreasing size before accessing data[s.size] can cause s.size to be negative if stack is empty.

2. Step 2: Understand runtime error cause

   Accessing data with negative index causes runtime panic (index out of range).

3. Final Answer:

   Decreasing size before checking if stack is empty causes negative index. -> Option A

4. Quick Check:

   Pop decrements size before check = negative index [OK]

Hint: Check empty before decrementing size [OK]

Common Mistakes:

• Not checking if stack is empty before pop
• Resizing array incorrectly
• Accessing data after size decrement without validation

5. You want to implement a dynamic stack in Go that shrinks its underlying array when the number of elements falls below one-fourth of the capacity, but never shrinks below a minimum capacity of 4. Which code snippet correctly implements this logic inside the Pop method?

hard

A. if s.size < len(s.data)/2 && len(s.data)/4 >= 4 { newData := make([]int, len(s.data)/4); copy(newData, s.data[:s.size]); s.data = newData }

B. if s.size < len(s.data)/4 { s.data = s.data[:len(s.data)/2] }

C. if s.size < 4 { s.data = make([]int, 4) }

D. if s.size < len(s.data)/4 && len(s.data)/2 >= 4 { newData := make([]int, len(s.data)/2); copy(newData, s.data[:s.size]); s.data = newData }

Solution

1. Step 1: Understand shrinking condition

   Shrink when size < 1/4 capacity, but only if new capacity ≥ 4.

2. Step 2: Check each option

   if s.size < len(s.data)/4 && len(s.data)/2 >= 4 { newData := make([]int, len(s.data)/2); copy(newData, s.data[:s.size]); s.data = newData } checks size < 1/4 capacity and new capacity ≥ 4, then resizes to half capacity correctly.

3. Final Answer:

   if s.size < len(s.data)/4 && len(s.data)/2 >= 4 { newData := make([]int, len(s.data)/2); copy(newData, s.data[:s.size]); s.data = newData } -> Option D

4. Quick Check:

   Shrink only if size < 1/4 and new capacity ≥ 4 [OK]

Hint: Shrink only if new capacity >= minimum size [OK]

Common Mistakes:

• Shrinking below minimum capacity
• Incorrect condition for shrinking
• Using wrong new capacity size