Heap Extract Min or Max Bubble Down in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to get the index of the left child in a heap.

DSA Go

func leftChild(index int) int {
    return 2*index [1] 1
}

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

A*

B-

C+

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of addition.

Using multiplication or division incorrectly.

In a heap, the left child of a node at index i is at 2*i + 1.

2fill in blank

medium

Complete the code to swap two elements in the heap slice.

DSA Go

func swap(heap []int, i, j int) {
    heap[i], heap[j] = heap[j], heap[1]i
}

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

A*

B(

C]

D[

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses instead of square brackets.

Using multiplication symbol by mistake.

To access an element in a slice, use square brackets [].

3fill in blank

hard

Fix the error in the bubbleDown function to correctly compare and swap with the smaller child.

DSA Go

func bubbleDown(heap []int, index int) {
    smallest := index
    left := 2*index + 1
    right := 2*index + 2

    if left < len(heap) && heap[left] < heap[1]smallest {
        smallest = left
    }
    if right < len(heap) && heap[right] < heap[smallest] {
        smallest = right
    }

    if smallest != index {
        heap[index], heap[smallest] = heap[smallest], heap[index]
        bubbleDown(heap, smallest)
    }
}

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

A]

B[

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses instead of square brackets.

Using arithmetic operators instead of indexing.

To access an element in a slice, use square brackets [].

4fill in blank

hard

Fill both blanks to complete the extractMin function that removes the smallest element and restores the heap.

DSA Go

func extractMin(heap []int) (int, []int) {
    if len(heap) == 0 {
        return -1, heap
    }
    min := heap[0]
    heap[0] = heap[len(heap) [1] 1]
    heap = heap[:len(heap) [2] 1]
    bubbleDown(heap, 0)
    return min, heap
}

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

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using addition instead of subtraction.

Using multiplication or division incorrectly.

To remove the last element, subtract 1 from length for indexing and slicing.

5fill in blank

hard

Fill all three blanks to complete the bubbleDown function for a max heap.

DSA Go

func bubbleDownMax(heap []int, index int) {
    largest := index
    left := 2*index + 1
    right := 2*index + 2

    if left < len(heap) && heap[left] [1] heap[largest] {
        largest = left
    }
    if right < len(heap) && heap[right] [2] heap[largest] {
        largest = right
    }

    if largest != index {
        heap[index], heap[largest] = heap[largest], heap[index]
        bubbleDownMax(heap, [3])
    }
}

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

A>

B<

Clargest

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using less than instead of greater than for comparisons.

Recursing on the wrong index.

For max heap, swap with child if child is greater; recurse on largest index.

Practice

1. What is the main purpose of the bubble down operation in a min-heap after extracting the minimum element?

easy

A. To add a new element at the end of the heap

B. To restore the heap property by moving the root element down to its correct position

C. To swap the minimum element with the maximum element

D. To remove all leaf nodes from the heap

Solution

1. Step 1: Understand extract-min operation

   When the minimum element (root) is removed, the last element is moved to the root.

2. Step 2: Apply bubble down to restore heap

   The bubble down moves this element down the tree to restore the min-heap property.

3. Final Answer:

   To restore the heap property by moving the root element down to its correct position -> Option B

4. Quick Check:

   Bubble down fixes heap after extract-min [OK]

Hint: Bubble down fixes heap from root after removal [OK]

Common Mistakes:

• Thinking bubble down adds elements
• Confusing bubble down with bubble up
• Believing bubble down swaps min and max

2. Which of the following Go code snippets correctly performs the bubble down operation for a min-heap starting at index i in slice heap?

easy

A. for {
left := 2*i + 1
right := 2*i + 2
smallest := i
if left < len(heap) && heap[left] < heap[smallest] {
smallest = left
}
if right < len(heap) && heap[right] < heap[smallest] {
smallest = right
}
if smallest == i {
break
}
heap[i], heap[smallest] = heap[smallest], heap[i]
i = smallest
}

B. for i < len(heap) {
left := 2*i
right := 2*i + 1
if heap[left] > heap[right] {
heap[i], heap[left] = heap[left], heap[i]
i = left
} else {
heap[i], heap[right] = heap[right], heap[i]
i = right
}
}

C. for i > 0 {
parent := (i - 1) / 2
if heap[i] < heap[parent] {
heap[i], heap[parent] = heap[parent], heap[i]
i = parent
} else {
break
}
}

D. for {
left := 2*i + 1
right := 2*i + 2
largest := i
if left < len(heap) && heap[left] > heap[largest] {
largest = left
}
if right < len(heap) && heap[right] > heap[largest] {
largest = right
}
if largest == i {
break
}
heap[i], heap[largest] = heap[largest], heap[i]
i = largest
}

Solution

1. Step 1: Identify bubble down for min-heap

   Bubble down compares current node with children and swaps with smallest child.

2. Step 2: Check code correctness

   for { left := 2*i + 1 right := 2*i + 2 smallest := i if left < len(heap) && heap[left] < heap[smallest] { smallest = left } if right < len(heap) && heap[right] < heap[smallest] { smallest = right } if smallest == i { break } heap[i], heap[smallest] = heap[smallest], heap[i] i = smallest } correctly finds smallest child and swaps until heap property is restored.

3. Final Answer:

   Option A code snippet correctly implements bubble down for min-heap -> Option A

4. Quick Check:

   Min-heap bubble down uses smallest child swap [OK]

Hint: Bubble down swaps with smallest child in min-heap [OK]

Common Mistakes:

• Using 1-based indexing instead of 0-based
• Swapping with largest child in min-heap
• Using bubble up logic instead of bubble down

3. Given the min-heap represented by the slice heap := []int{2, 4, 5, 10, 8, 15}, what will be the heap slice after extracting the minimum element and performing bubble down?

medium

A. [4, 8, 5, 10, 15]

B. [5, 4, 15, 10, 8]

C. [4, 10, 5, 15, 8]

D. [8, 4, 5, 10, 15]

Solution

1. Step 1: Remove min and replace root

   Remove 2 (root), move last element 15 to root: [15,4,5,10,8]

2. Step 2: Bubble down 15

   Compare 15 with children 4 and 5; swap with 4 (smallest). Now heap: [4,15,5,10,8]. Next, bubble down 15 at index 1: children 10 and 8; swap with 8 (smallest). Final heap: [4,8,5,10,15]

3. Final Answer:

   [4, 8, 5, 10, 15] -> Option A

4. Quick Check:

   Extract min + bubble down = [4,8,5,10,15] [OK]

Hint: Replace root with last, then swap down with smaller child [OK]

Common Mistakes:

• Not swapping with smallest child first
• Forgetting to move last element to root
• Swapping with larger child

4. Consider the following Go code snippet for bubble down in a max-heap. Identify the bug that causes incorrect heap after extraction:

func bubbleDown(heap []int, i int) {
  for {
    left := 2*i + 1
    right := 2*i + 2
    largest := i
    if left < len(heap) && heap[left] < heap[largest] {
      largest = left
    }
    if right < len(heap) && heap[right] < heap[largest] {
      largest = right
    }
    if largest == i {
      break
    }
    heap[i], heap[largest] = heap[largest], heap[i]
    i = largest
  }
}

medium

A. The loop condition should be 'while i > 0' instead of infinite loop

B. The indices for left and right children are incorrect

C. Swapping elements should be done after the loop ends

D. The comparisons use '<' instead of '>' for max-heap

Solution

1. Step 1: Understand max-heap bubble down logic

   In max-heap, bubble down swaps with the largest child, so comparisons should find the largest value.

2. Step 2: Identify incorrect comparison operators

   The code uses '<' to find smaller child, which is wrong for max-heap; it should use '>' to find largest child.

3. Final Answer:

   The comparisons use '<' instead of '>' for max-heap -> Option D

4. Quick Check:

   Max-heap bubble down uses '>' comparisons [OK]

Hint: Max-heap bubble down compares with '>' not '<' [OK]

Common Mistakes:

• Using wrong comparison operators for heap type
• Incorrect child index calculations
• Misplacing swap inside or outside loop

5. You have a max-heap represented as heap := []int{50, 30, 40, 10, 20, 35}. After extracting the maximum element and performing bubble down, what will be the resulting heap slice?

hard

A. [35, 30, 40, 10, 20]

B. [30, 20, 35, 10, 40]

C. [40, 30, 35, 10, 20]

D. [40, 30, 35, 20, 10]

Solution

1. Step 1: Remove max and replace root

   Remove 50 (root), move last element 35 to root: [35,30,40,10,20]

2. Step 2: Bubble down 35

   Compare 35 with children 30 and 40; swap with 40 (largest). Now heap: [40,30,35,10,20]. Next, bubble down 35 at index 2: children indices 5 and 6 do not exist (len=5), so stop. This restores the max-heap property.

3. Final Answer:

   [40, 30, 35, 10, 20] -> Option C

4. Quick Check:

   Extract max + bubble down = [40,30,35,10,20] [OK]

Hint: Swap with largest child repeatedly until heap property restored [OK]

Common Mistakes:

• Not swapping with largest child first
• Stopping bubble down too early
• Ignoring heap property for children