Heap Extract Min or Max Bubble Down in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Extract-Min on a Min-Heap

What is the state of the heap array after extracting the minimum element once from this min-heap?

Initial heap array (index 0 is root): [2, 5, 3, 8, 10, 15]

DSA Go

heap := []int{2, 5, 3, 8, 10, 15}
// Extract min operation
min := heap[0]
heap[0] = heap[len(heap)-1]
heap = heap[:len(heap)-1]
// Bubble down
index := 0
for {
  left := 2*index + 1
  right := 2*index + 2
  smallest := index
  if left < len(heap) && heap[left] < heap[smallest] {
    smallest = left
  }
  if right < len(heap) && heap[right] < heap[smallest] {
    smallest = right
  }
  if smallest == index {
    break
  }
  heap[index], heap[smallest] = heap[smallest], heap[index]
  index = smallest
}
fmt.Println(heap)

A [3, 5, 15, 8, 10]

B [5, 8, 3, 10, 15]

C [3, 5, 10, 8, 15]

D [5, 3, 8, 10, 15]

💡 Hint

Remember to replace the root with the last element and then bubble down by swapping with the smaller child.

After removing 2, the last element 15 moves to root. Then bubble down swaps 15 with 3 (smaller child), and stops because the new position (index 2) has no children (left child at index 5 >= length 5). Final heap is [3, 5, 15, 8, 10].

❓ Predict Output

intermediate

Output of Extract-Max on a Max-Heap

Given this max-heap array:

[20, 18, 15, 10, 12, 9, 8]

What is the heap array after extracting the maximum element once?

DSA Go

heap := []int{20, 18, 15, 10, 12, 9, 8}
max := heap[0]
heap[0] = heap[len(heap)-1]
heap = heap[:len(heap)-1]
index := 0
for {
  left := 2*index + 1
  right := 2*index + 2
  largest := index
  if left < len(heap) && heap[left] > heap[largest] {
    largest = left
  }
  if right < len(heap) && heap[right] > heap[largest] {
    largest = right
  }
  if largest == index {
    break
  }
  heap[index], heap[largest] = heap[largest], heap[index]
  index = largest
}
fmt.Println(heap)

A [18, 15, 9, 10, 12, 8]

B [18, 12, 9, 10, 8, 15]

C [15, 18, 9, 10, 12, 8]

D [18, 12, 15, 10, 8, 9]

💡 Hint

After replacing root with last element, bubble down by swapping with the larger child.

After removing 20, 8 moves to root. It swaps with 18 (larger child), then at index 1 (now 8) swaps with 12 (larger child there), then stops. Final heap is [18, 12, 15, 10, 8, 9].

🔧 Debug

advanced

Identify the Bug in Bubble Down for Min-Heap

This Go code tries to bubble down after extract-min on a min-heap. What error will it cause when run?

DSA Go

heap := []int{1, 3, 6, 5, 9, 8}
heap[0] = heap[len(heap)-1]
heap = heap[:len(heap)-1]
index := 0
for {
  left := 2*index + 1
  right := 2*index + 2
  smallest := index
  if left < len(heap) && heap[left] > heap[smallest] {
    smallest = left
  }
  if right < len(heap) && heap[right] < heap[smallest] {
    smallest = right
  }
  if smallest == index {
    break
  }
  heap[index], heap[smallest] = heap[smallest], heap[index]
  index = smallest
}
fmt.Println(heap)

A No error, prints [3, 5, 6, 8, 9]

B Index out of range runtime error

C Logic error: heap property violated, wrong output

D Compilation error due to invalid syntax

💡 Hint

Check the comparison signs in the if conditions for choosing smallest child.

The first if uses '>' instead of '<' causing wrong child selection. This breaks min-heap property but no runtime error occurs.

🧠 Conceptual

advanced

Why Bubble Down is Needed After Extract-Min or Extract-Max?

After removing the root element from a heap and replacing it with the last element, why do we need to bubble down?

A To restore the heap property by moving the new root down to its correct position

B To sort the entire heap array in ascending order

C To remove duplicate elements from the heap

D To increase the size of the heap by adding new elements

💡 Hint

Think about what happens to the heap property after replacing the root.

Replacing the root with the last element may break the heap property. Bubble down moves this element down to restore the heap order.

❓ Predict Output

expert

Final Heap After Multiple Extract-Min Operations

Starting with this min-heap array:

[4, 7, 6, 10, 9, 15, 20]

Perform extract-min operation twice in a row. What is the heap array after these two extractions?

DSA Go

heap := []int{4, 7, 6, 10, 9, 15, 20}
for i := 0; i < 2; i++ {
  heap[0] = heap[len(heap)-1]
  heap = heap[:len(heap)-1]
  index := 0
  for {
    left := 2*index + 1
    right := 2*index + 2
    smallest := index
    if left < len(heap) && heap[left] < heap[smallest] {
      smallest = left
    }
    if right < len(heap) && heap[right] < heap[smallest] {
      smallest = right
    }
    if smallest == index {
      break
    }
    heap[index], heap[smallest] = heap[smallest], heap[index]
    index = smallest
  }
}
fmt.Println(heap)

A [6, 9, 15, 10, 20]

B [7, 9, 15, 10, 20]

C [6, 9, 20, 10, 15]

D [7, 10, 15, 9, 20]

💡 Hint

Perform extract-min twice carefully, bubbling down each time.

First extraction removes 4, heap becomes [6,7,15,10,9,20] after bubble down. Second extraction removes 6, heap becomes [7,9,15,10,20].