Build Heap from Array Heapify in DSA C++ - Execution Trace

Concept Flow - Build Heap from Array Heapify

Start with input array

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Identify last non-leaf node

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For each node from last non-leaf to root

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Perform heapify on current node

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Heapify: Compare node with children

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Swap with largest child if needed

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Repeat heapify on swapped child

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Move to previous node

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All nodes heapified -> Max-Heap built

Start from the last non-leaf node and heapify each node upwards to the root to build a max-heap from the array.

Execution Sample

DSA C++

void buildHeap(int arr[], int n) {
  for (int i = n/2 - 1; i >= 0; i--) {
    heapify(arr, n, i);
  }
}

This code builds a max-heap from an unsorted array by calling heapify from the last non-leaf node up to the root.

Execution Table

Step	Operation	Node Index	Nodes in Array	Pointer Changes	Visual State
1	Start buildHeap	-	[4, 10, 3, 5, 1]	Set i = 1 (n/2 -1)	[4, 10, 3, 5, 1]
2	Heapify at index 1	1	[4, 10, 3, 5, 1]	Compare arr[1]=10 with children arr[3]=5, arr[4]=1	[4, 10, 3, 5, 1]
3	Heapify at index 1	1	[4, 10, 3, 5, 1]	No swap needed (10 is largest)	[4, 10, 3, 5, 1]
4	Move to index 0	0	[4, 10, 3, 5, 1]	Compare arr[0]=4 with children arr[1]=10, arr[2]=3	[4, 10, 3, 5, 1]
5	Swap arr[0] and arr[1]	0->1	[4, 10, 3, 5, 1]	Swap 4 and 10	[10, 4, 3, 5, 1]
6	Heapify at index 1	1	[10, 4, 3, 5, 1]	Compare arr[1]=4 with children arr[3]=5, arr[4]=1	[10, 4, 3, 5, 1]
7	Swap arr[1] and arr[3]	1->3	[10, 4, 3, 5, 1]	Swap 4 and 5	[10, 5, 3, 4, 1]
8	Heapify at index 3	3	[10, 5, 3, 4, 1]	No children, stop	[10, 5, 3, 4, 1]
9	All nodes heapified	-	[10, 5, 3, 4, 1]	Heap built	[10, 5, 3, 4, 1]

💡 All non-leaf nodes heapified, max-heap property satisfied for entire array.

Variable Tracker

Variable	Start	After Step 2	After Step 5	After Step 7	Final
i	-	1	0	0	-
arr	[4, 10, 3, 5, 1]	[4, 10, 3, 5, 1]	[10, 4, 3, 5, 1]	[10, 5, 3, 4, 1]	[10, 5, 3, 4, 1]
heapify node	-	1	1	3	-

Key Moments - 3 Insights

Why do we start heapify from index n/2 - 1 and not from the last element?

Because nodes from n/2 to n-1 are leaf nodes and already satisfy heap property. See execution_table step 1 where i starts at 1 for n=5.

Why do we heapify again at the child index after swapping?

Swapping may break heap property at the child, so we heapify recursively down. See steps 6-8 where after swapping at index 1, heapify continues at index 3.

What happens if no swap is needed during heapify?

Heapify stops for that node as heap property is satisfied. See step 3 where no swap occurs at index 1.

Visual Quiz - 3 Questions

Look at the execution_table, what is the array state after step 5?

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

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

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

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

💡 Hint

Check the Visual State column at step 5 in execution_table.

At which step does the heapify process stop for the node at index 3?

A Step 6

B Step 8

C Step 7

D Step 9

💡 Hint

Look for 'No children, stop' in the Action column in execution_table.

If the input array was already a max-heap, how would the execution_table change?

A More swaps would occur

B Heapify would start from index 0 instead of n/2 -1

C No swaps would occur during heapify

D Heapify would not be called at all

💡 Hint

Refer to step 3 where no swap is needed when heap property is already satisfied.

Concept Snapshot

Build Heap from Array (Heapify):
- Start from last non-leaf node (n/2 -1) down to root
- For each node, heapify: compare with children
- Swap with largest child if needed
- Repeat heapify down the subtree
- Result: array satisfies max-heap property

Full Transcript

To build a max-heap from an array, we start heapifying from the last non-leaf node, which is at index n/2 -1, moving upwards to the root. Heapify compares the node with its children and swaps with the largest child if needed, then continues heapifying down the subtree. This process ensures the max-heap property is satisfied for the entire array. Leaf nodes do not need heapify. The example array [4, 10, 3, 5, 1] is transformed step-by-step into [10, 5, 3, 4, 1].