Build Heap from Array Heapify in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Build Max-Heap from Array

What is the printed state of the array after building a max-heap using heapify on the input array?

DSA Go

package main

import "fmt"

func heapify(arr []int, n int, i int) {
    largest := i
    left := 2*i + 1
    right := 2*i + 2

    if left < n && arr[left] > arr[largest] {
        largest = left
    }

    if right < n && arr[right] > arr[largest] {
        largest = right
    }

    if largest != i {
        arr[i], arr[largest] = arr[largest], arr[i]
        heapify(arr, n, largest)
    }
}

func buildMaxHeap(arr []int) {
    n := len(arr)
    for i := n/2 - 1; i >= 0; i-- {
        heapify(arr, n, i)
    }
}

func main() {
    arr := []int{3, 9, 2, 1, 4, 5}
    buildMaxHeap(arr)
    fmt.Println(arr)
}

A [5, 4, 3, 1, 9, 2]

B [3, 9, 5, 1, 4, 2]

C [9, 3, 5, 1, 4, 2]

D [9, 4, 5, 1, 3, 2]

💡 Hint

Remember that buildMaxHeap starts heapifying from the last non-leaf node upwards.

The buildMaxHeap function heapifies from the middle of the array towards the root, ensuring the max-heap property. After heapify, the largest element is at the root, and the array is rearranged accordingly.

❓ Predict Output

intermediate

Output of Build Min-Heap from Array

What is the printed state of the array after building a min-heap using heapify on the input array?

DSA Go

package main

import "fmt"

func heapify(arr []int, n int, i int) {
    smallest := i
    left := 2*i + 1
    right := 2*i + 2

    if left < n && arr[left] < arr[smallest] {
        smallest = left
    }

    if right < n && arr[right] < arr[smallest] {
        smallest = right
    }

    if smallest != i {
        arr[i], arr[smallest] = arr[smallest], arr[i]
        heapify(arr, n, smallest)
    }
}

func buildMinHeap(arr []int) {
    n := len(arr)
    for i := n/2 - 1; i >= 0; i-- {
        heapify(arr, n, i)
    }
}

func main() {
    arr := []int{7, 2, 6, 3, 9, 1}
    buildMinHeap(arr)
    fmt.Println(arr)
}

A [1, 2, 3, 7, 9, 6]

B [1, 2, 6, 3, 9, 7]

C [2, 3, 1, 7, 9, 6]

D [1, 3, 2, 6, 7, 9]

💡 Hint

Build min-heap by heapifying from last non-leaf node upwards, ensuring smallest element at root.

The buildMinHeap function heapifies the array so that the smallest element is at the root and the min-heap property is maintained throughout the tree.

🔧 Debug

advanced

Identify the Bug in Heapify Function

What error will occur when running this heapify function on an array?

DSA Go

func heapify(arr []int, n int, i int) {
    largest := i
    left := 2*i
    right := 2*i + 1

    if left < n && arr[left] > arr[largest] {
        largest = left
    }

    if right < n && arr[right] > arr[largest] {
        largest = right
    }

    if largest != i {
        arr[i], arr[largest] = arr[largest], arr[i]
        heapify(arr, n, largest)
    }
}

A Index out of range runtime panic

B No error, runs correctly

C Compilation error due to missing return

D Stack overflow due to infinite recursion

💡 Hint

Check how left and right child indices are calculated for a zero-based array.

The left and right child indices are incorrectly calculated as 2*i and 2*i+1 instead of 2*i+1 and 2*i+2, causing out-of-range access for i=0.

❓ Predict Output

advanced

Resulting Array After Partial Heapify

Given the array and a call to heapify at index 1, what is the resulting array?

DSA Go

package main

import "fmt"

func heapify(arr []int, n int, i int) {
    largest := i
    left := 2*i + 1
    right := 2*i + 2

    if left < n && arr[left] > arr[largest] {
        largest = left
    }

    if right < n && arr[right] > arr[largest] {
        largest = right
    }

    if largest != i {
        arr[i], arr[largest] = arr[largest], arr[i]
        heapify(arr, n, largest)
    }
}

func main() {
    arr := []int{4, 10, 3, 5, 1}
    heapify(arr, len(arr), 1)
    fmt.Println(arr)
}

A [4, 10, 3, 5, 1]

B ]1 ,5 ,3 ,01 ,4[

C 4, 10, 3, 5, 1]

D [4, 10, 3, 5, 1

💡 Hint

Check if the subtree rooted at index 1 needs rearrangement.

At index 1, the value is 10, children are 5 and 1 which are smaller, so no swaps happen. The array remains unchanged.

❓ Predict Output

expert

Final Array After Build Max-Heap on Large Input

What is the final array after building a max-heap from this input array?

DSA Go

package main

import "fmt"

func heapify(arr []int, n int, i int) {
    largest := i
    left := 2*i + 1
    right := 2*i + 2

    if left < n && arr[left] > arr[largest] {
        largest = left
    }

    if right < n && arr[right] > arr[largest] {
        largest = right
    }

    if largest != i {
        arr[i], arr[largest] = arr[largest], arr[i]
        heapify(arr, n, largest)
    }
}

func buildMaxHeap(arr []int) {
    n := len(arr)
    for i := n/2 - 1; i >= 0; i-- {
        heapify(arr, n, i)
    }
}

func main() {
    arr := []int{12, 11, 13, 5, 6, 7, 17, 1, 8, 9}
    buildMaxHeap(arr)
    fmt.Println(arr)
}

A [13, 11, 17, 5, 6, 7, 12, 1, 8, 9]

B [17, 11, 13, 9, 6, 7, 12, 1, 8, 10]

C [17, 11, 13, 9, 6, 7, 12, 1, 8, 5]

D [17, 13, 12, 9, 11, 7, 6, 1, 8, 5]

💡 Hint

Build max-heap by heapifying from last non-leaf node upwards.

After heapify, the largest element 17 is at root, and the array is rearranged to maintain max-heap property.

Practice

1. What is the main purpose of the heapify process when building a heap from an array?

easy

A. To rearrange the array elements to satisfy the heap property

B. To sort the array in ascending order

C. To find the maximum element in the array

D. To reverse the array elements

Solution

1. Step 1: Understand heap property

   A heap is a special tree-based structure where each parent node is greater (max-heap) or smaller (min-heap) than its children.

2. Step 2: Role of heapify

   The heapify process rearranges elements to ensure this property holds for every subtree starting from a given node.

3. Final Answer:

   To rearrange the array elements to satisfy the heap property -> Option A

4. Quick Check:

   Heapify = Rearrange for heap property [OK]

Hint: Heapify fixes subtree to maintain heap order [OK]

Common Mistakes:

• Confusing heapify with sorting
• Thinking heapify finds max element only
• Assuming heapify reverses array

2. Which of the following is the correct way to call heapify on a node index i in Go when building a max heap from an array arr of size n?

easy

A. heapify(n, arr, i)

B. heapify(i, arr, n)

C. heapify(arr, i, n)

D. heapify(arr, n, i)

Solution

1. Step 1: Understand heapify function parameters

   Heapify usually takes the array, size of heap, and the index to heapify as parameters in that order.

2. Step 2: Match parameter order

   heapify(arr, n, i) matches the correct order: array, size, index.

3. Final Answer:

   heapify(arr, n, i) -> Option D

4. Quick Check:

   heapify(arr, n, i) is correct call [OK]

Hint: Remember: heapify(array, size, index) [OK]

Common Mistakes:

• Mixing parameter order
• Passing index before array
• Confusing size and index positions

3. Given the array [3, 9, 2, 1, 4, 5], what is the array after building a max heap using heapify from the bottom up?

medium

A. [5, 4, 9, 1, 3, 2]

B. [9, 4, 5, 3, 1, 2]

C. [9, 4, 5, 1, 3, 2]

D. [9, 3, 5, 1, 4, 2]

Solution

1. Step 1: Start heapify from last non-leaf node

   Last non-leaf node index = n/2 -1 = 6/2 -1 = 2. Heapify nodes 2, 1, 0 in order.

2. Step 2: Heapify index 2

   Node 2 value=2, children 5 (index 5). Swap 2 and 5. Array becomes [3, 9, 5, 1, 4, 2].

3. Step 3: Heapify index 1

   Node 1 value=9, children 1 and 4. 9 is largest, no change.

4. Step 4: Heapify index 0

   Node 0 value=3, children 9 and 5. Swap 3 and 9. Array becomes [9, 3, 5, 1, 4, 2]. Heapify index 1 again: 3 with children 1 and 4, swap 3 and 4. Final array: [9, 4, 5, 1, 3, 2].

5. Final Answer:

   [9, 4, 5, 1, 3, 2] -> Option C

6. Quick Check:

   Bottom-up heapify gives [9,4,5,1,3,2] [OK]

Hint: Heapify bottom-up swaps to fix largest child first [OK]

Common Mistakes:

• Not heapifying children after swap
• Starting heapify from wrong index
• Confusing max-heap with min-heap

4. Consider this Go code snippet to build a max heap from array arr. What is the error?

func buildHeap(arr []int) {
    n := len(arr)
    for i := n / 2; i >= 0; i-- {
        heapify(arr, n, i)
    }
}

medium

A. heapify function is missing the array size parameter

B. The loop should start from n/2 - 1, not n/2

C. The loop should increment i, not decrement

D. The loop condition should be i > 0, not i >= 0

Solution

1. Step 1: Identify correct start index for heapify

   Heapify starts from last non-leaf node at index n/2 - 1, not n/2.

2. Step 2: Analyze loop range

   Starting at n/2 includes one extra index which is a leaf node, unnecessary and may cause incorrect behavior.

3. Final Answer:

   The loop should start from n/2 - 1, not n/2 -> Option B

4. Quick Check:

   Start heapify at n/2 - 1 [OK]

Hint: Start heapify at last non-leaf node: n/2 - 1 [OK]

Common Mistakes:

• Starting loop at n/2 instead of n/2 - 1
• Loop increment instead of decrement
• Wrong loop condition excluding index 0

5. You have an array [4, 10, 3, 5, 1]. After building a max heap using heapify, you want to insert a new element 15 and maintain the max heap property. Which sequence of operations correctly maintains the heap?

hard

A. Append 15, then bubble up the new element until heap property is restored

B. Append 15, then heapify only the root node

C. Append 15, then heapify from last non-leaf node down to root

D. Replace root with 15, then heapify entire array

Solution

1. Step 1: Understand insertion in heap

   To insert, add element at end, then restore heap by bubbling up (swapping with parent if larger).

2. Step 2: Why bubble up?

   Heapify from bottom up is for building heap initially; for insertion, bubbling up fixes heap efficiently.

3. Final Answer:

   Append 15, then bubble up the new element until heap property is restored -> Option A

4. Quick Check:

   Insert + bubble up maintains heap [OK]

Hint: Insert at end, bubble up to fix heap [OK]

Common Mistakes:

• Using heapify from root after insertion
• Heapifying only root node
• Replacing root without adjusting heap