Build Heap from Array Heapify in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

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

Heapify rearranges elements to satisfy the heap property, ensuring each parent node is greater (max-heap) or smaller (min-heap) than its children.

beginner

In a zero-indexed array, what is the formula to find the left child of a node at index i?

Left child index = 2 * i + 1

intermediate

Why do we start heapify from the last non-leaf node when building a heap from an array?

Because leaf nodes are already heaps, starting from the last non-leaf node ensures all subtrees satisfy the heap property when moving upward.

intermediate

What is the time complexity of building a heap from an unsorted array using heapify?

O(n), where n is the number of elements in the array.

beginner

Explain the difference between max-heap and min-heap in the context of heapify.

Max-heap ensures each parent node is greater than its children, while min-heap ensures each parent node is smaller than its children during heapify.

What is the index of the last non-leaf node in a heap of size n (0-indexed)?

A n/3

B n - 1

C n/2

D n/2 - 1

The last non-leaf node is at index n/2 - 1 because nodes after this are leaves.

Which of the following best describes the heapify operation?

A Rearranging a subtree to satisfy the heap property

B Sorting the entire array

C Inserting a new element into the heap

D Removing the root element

Heapify fixes the heap property in a subtree rooted at a given node.

What is the time complexity of heapify on a single node?

A O(log n)

B O(n)

C O(1)

D O(n log n)

Heapify on one node takes O(log n) time because it may move down the tree height.

When building a heap from an array, why is the overall time complexity O(n) and not O(n log n)?

A Because only half the nodes are heapified

B Because heapify is called fewer times on nodes near the bottom

C Because heapify is O(1) for all nodes

D Because the array is already sorted

Nodes near the bottom have smaller heights, so heapify is faster there, making total O(n).

In a max-heap, after heapify, what is true about the root node?

A It is the average of all elements

B It is the smallest element in the heap

C It is the largest element in the heap

D It is the last inserted element

Max-heap property ensures the root is the largest element.

Describe the step-by-step process of building a max-heap from an unsorted array using heapify.

Think about fixing subtrees from bottom to top.

Explain why heapify is efficient when building a heap compared to inserting elements one by one.

Compare bulk building vs incremental insertion.

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 reverse the array elements

D. To remove duplicate elements from the array

Solution

1. Step 1: Understand heap property

   The heap property requires each parent node to be greater (max-heap) or smaller (min-heap) than its children.

2. Step 2: Role of heapify

   Heapify rearranges elements to ensure this property holds for every subtree starting from a 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:

• Thinking heapify sorts the array
• Confusing heapify with reversing array
• Assuming heapify removes duplicates

2. Which of the following is the correct Python syntax to call the heapify function from the heapq module on a list named arr?

easy

A. arr.heapify()

B. heapify.heapq(arr)

C. heapq.heapify(arr)

D. heapq.heapify = arr

Solution

1. Step 1: Identify module and function

   The heapq module provides the heapify function to convert a list into a heap.

2. Step 2: Correct function call syntax

   The correct syntax is heapq.heapify(arr) after importing heapq.

3. Final Answer:

   heapq.heapify(arr) -> Option C

4. Quick Check:

   heapq.heapify(list) is correct call [OK]

Hint: Call heapify as heapq.heapify(list) after import [OK]

Common Mistakes:

• Using wrong order like heapify.heapq(arr)
• Calling heapify as a method on list
• Assigning heapify function instead of calling

3. What will be the output of the following Python code?

import heapq
arr = [3, 1, 6, 5, 2, 4]
heapq.heapify(arr)
print(arr)

medium

A. [6, 5, 4, 3, 2, 1]

B. [1, 2, 4, 5, 3, 6]

C. [1, 3, 2, 5, 6, 4]

D. [3, 1, 6, 5, 2, 4]

Solution

1. Step 1: Understand heapq.heapify behavior

   heapq.heapify converts list into a min-heap, smallest element at root (index 0).

2. Step 2: Apply heapify on given list

   After heapify, the list becomes [1, 2, 4, 5, 3, 6], which satisfies min-heap property.

3. Final Answer:

   [1, 2, 4, 5, 3, 6] -> Option B

4. Quick Check:

   heapify(min-heap) = [1, 2, 4, 5, 3, 6] [OK]

Hint: heapq.heapify creates min-heap, smallest first [OK]

Common Mistakes:

• Expecting sorted list instead of heap order
• Confusing max-heap with min-heap output
• Assuming original list order remains

4. The following code attempts to build a heap from an array but does not produce the correct heap. What is the error?

def build_heap(arr):
    n = len(arr)
    for i in range(n):
        heapify(arr, i, n)

def heapify(arr, i, n):
    largest = i
    left = 2 * i + 1
    right = 2 * i + 2
    if left < n and arr[left] > arr[largest]:
        largest = left
    if right < n and arr[right] > arr[largest]:
        largest = right
    if largest != i:
        arr[i], arr[largest] = arr[largest], arr[i]
        heapify(arr, largest, n)

arr = [4, 10, 3, 5, 1]
build_heap(arr)
print(arr)

medium

A. The loop should run from n//2 - 1 down to 0, not from 0 to n

B. The heapify function should use min-heap logic instead of max-heap

C. The swap inside heapify is missing

D. The array length n is calculated incorrectly

Solution

1. Step 1: Analyze build_heap loop range

   Heap building requires heapify calls from last non-leaf node down to root, i.e., from n//2 - 1 down to 0.

2. Step 2: Identify incorrect loop direction

   The code loops from 0 to n, which is incorrect and misses proper heapify order.

3. Final Answer:

   The loop should run from n//2 - 1 down to 0, not from 0 to n -> Option A

4. Quick Check:

   Build heap loop must go bottom-up [OK]

Hint: Build heap bottom-up from last parent to root [OK]

Common Mistakes:

• Looping top-down instead of bottom-up
• Confusing min-heap and max-heap logic
• Forgetting to swap elements in heapify

5. Given an array arr = [9, 4, 7, 1, -2, 6, 5], you want to build a max-heap using the heapify method. Which sequence of indices should you call heapify on to build the heap correctly?

hard

A. From index 0 to len(arr)-1 in ascending order

B. Only on index 0

C. From index len(arr)-1 down to 0 in descending order

D. From index len(arr)//2 - 1 down to 0 in descending order

Solution

1. Step 1: Identify last non-leaf node

   For array length 7, last non-leaf node is at index 7//2 - 1 = 2.

2. Step 2: Correct heapify call order

   Heapify must be called from index 2 down to 0 to ensure all subtrees satisfy max-heap property.

3. Final Answer:

   From index len(arr)//2 - 1 down to 0 in descending order -> Option D

4. Quick Check:

   Heapify bottom-up from last parent to root [OK]

Hint: Start heapify from last parent index down to root [OK]

Common Mistakes:

• Calling heapify in ascending order
• Calling heapify only on root
• Calling heapify on all indices including leaves