Heapify — Building a Heap Efficiently

Consider this scenario: you're building a task scheduling system for a data center with thousands of servers. Each server submits tasks with priority levels, and your system must process them in priority order. The catch? At system initialization, you receive a massive dump of 10 million pending tasks from a crash recovery log—all at once, in arbitrary order.

You know heaps are ideal for priority processing. But how do you convert those 10 million unsorted task records into a valid heap? Do you insert them one by one? Is there a faster way?

This question—how to efficiently build a heap from an existing collection—is one of the most important problems in heap algorithms. The naive approach seems obvious, but as we'll discover, it's not optimal. The optimal approach, called heapify (or build-heap), achieves something remarkable: it constructs a valid heap in O(n) time, not O(n log n). Understanding why this works requires grasping some beautiful mathematical insights about tree structure.

Let's define the heap construction problem with mathematical precision. Ambiguity in problem statements leads to ambiguity in solutions, so we'll be rigorous.

Input:

An array A of n comparable elements in arbitrary (unsorted) order:

A = [a₀, a₁, a₂, ..., aₙ₋₁]

Output:

Rearrange the elements of A in-place such that A satisfies the heap property.

The Max-Heap Property (formal):

For all indices i where i > 0:

A[parent(i)] ≥ A[i]

where parent(i) = ⌊(i - 1) / 2⌋.

Equivalently: every node's value is at least as large as the values of its children.

The Min-Heap Property (formal):

For all indices i where i > 0:

A[parent(i)] ≤ A[i]

Equivalently: every node's value is at most as small as the values of its children.

What We're NOT Doing:

It's important to distinguish heap construction from related operations:

1. Not sorting — We're not producing a sorted array. A heap is only partially ordered. The heap property is weaker than the sorted property.

2. Not creating a new heap and inserting — While this works, it's not in-place (requires new allocation) and, as we'll see, it's slower than necessary.

3. Not building from a data stream — We have all elements available upfront. This batch availability is what enables the O(n) algorithm.

Example:

Input Array:  [3, 9, 2, 1, 4, 5]

Valid Max-Heap Output: [9, 4, 5, 1, 3, 2]

Verification:
- parent(1)=0: A[0]=9 ≥ A[1]=4 ✓
- parent(2)=0: A[0]=9 ≥ A[2]=5 ✓
- parent(3)=1: A[1]=4 ≥ A[3]=1 ✓
- parent(4)=1: A[1]=4 ≥ A[4]=3 ✓
- parent(5)=2: A[2]=5 ≥ A[5]=2 ✓

Note that there are multiple valid heap configurations for the same input. Our algorithm needs to produce any valid heap, not a specific one.

The first solution most developers think of is straightforward: create an empty heap, then insert elements one by one.

Naive Algorithm:

BUILD-HEAP-NAIVE(A):
    H = new empty heap
    for each element x in A:
        H.insert(x)
    return H

This works correctly—it definitely produces a valid heap. But let's analyze its complexity.

Time Complexity Analysis:

Each insert operation is O(log k) where k is the current heap size. For the entire sequence:

Total Work:

T(n) = Σ(i=1 to n) log i = log(1) + log(2) + log(3) + ... + log(n)
     = log(1 × 2 × 3 × ... × n)
     = log(n!)

Using Stirling's approximation: log(n!) ≈ n log(n) - n log(e) = Θ(n log n)

Therefore, the naive approach is O(n log n).

Additional Overhead:

Beyond the worse complexity, the naive approach has practical drawbacks:

1. Memory allocation — Creating a new heap requires allocating n elements of space, violating the in-place constraint.

2. Cache inefficiency — Growing arrays cause cache misses; the final elements are far from the first in memory.

3. More swaps — Each insertion potentially swaps all the way to the root, performing ~n log n swaps total.

Before diving into the optimal algorithm, let's understand why the difference between O(n log n) and O(n) actually matters in practice. These scenarios demonstrate that heap construction speed is not merely theoretical.

Scenario 1: HeapSort Algorithm

HeapSort is a comparison-based sorting algorithm that:

1. Builds a max-heap from the input array
2. Repeatedly extracts the maximum to the end of the array

The overall complexity is:

• Build heap: T_build
• Extract n times: O(n log n)
• Total: T_build + O(n log n)

If T_build = O(n log n), total is O(n log n) — optimal for comparison sorts. If T_build = O(n), total is still O(n log n) — but the constant factor improves!

For HeapSort to be competitive with QuickSort and MergeSort, the O(n) build is essential. With O(n log n) build, the constant factor makes HeapSort noticeably slower.

Scenario 2: Finding K Largest Elements

To find the K largest elements in an array of n elements:

Approach A: Build a max-heap of all n elements, extract K times.

• Time: O(n) + O(K log n)

Approach B: Maintain a min-heap of K elements, scanning through n.

• Time: O(n log K)

For small K, Approach A is better—but only if build-heap is O(n). With O(n log n) build, the advantage disappears.

Scenario 3: Batch Priority Queue Initialization

Database query planners, operating system schedulers, and network routers often initialize with a batch of existing work items:

• OS reboot: thousands of pending processes from hibernation state
• Database startup: queued transactions from write-ahead log
• Network router: buffered packets from interface reset

With millions of items, the difference between 20 million operations and 2 million operations means the difference between seconds of initialization and sub-second—critical for high-availability systems.

The optimal heap construction algorithm is based on a beautiful observation about tree structure. Let's build up to this insight.

Observation 1: A Single Node Is a Valid Heap

A heap with one element trivially satisfies the heap property—there are no parent-child pairs to violate it.

Observation 2: Leaves Have No Children

In a complete binary tree represented as an array:

• Nodes with indices 0 to ⌊n/2⌋ - 1 are internal nodes (have at least one child)
• Nodes with indices ⌊n/2⌋ to n - 1 are leaves (have no children)

This can be derived from the child index formulas:

• Left child of index i: 2i + 1
• For this to be valid: 2i + 1 < n, so i < (n-1)/2, meaning i ≤ ⌊(n-2)/2⌋ = ⌊n/2⌋ - 1

Observation 3: Every Leaf Is a Valid Heap

Since leaves have no children, there are no heap property constraints to satisfy. Every leaf, considered as a subtree of one node, is already a valid heap.

The Critical Insight:

In a complete binary tree with n nodes, approximately half the nodes are leaves. These leaves are already valid heaps without any work.

For n = 15 (a complete tree with 4 levels):

Level 0:     1 node  (root)           — internal
Level 1:     2 nodes                  — internal
Level 2:     4 nodes                  — internal
Level 3:     8 nodes                  — LEAVES

Total: 15 nodes
Leaves: 8 nodes (53%)
Internal: 7 nodes (47%)

For a general complete binary tree:

• Number of leaves: ⌈n/2⌉
• Number of internal nodes: ⌊n/2⌋

What This Means:

We don't need to do any work on half the array! The naive approach (repeated insertion) doesn't exploit this—it processes every element. But if we could somehow "merge" valid sub-heaps into larger valid heaps, we could skip the leaves entirely and work only on internal nodes.

This is exactly what the bottom-up heapify algorithm does.

To understand bottom-up heapify, we need to see how small heaps can combine into larger heaps. This compositional view is fundamental.

Composition Theorem:

Given:

• A value v at some position
• A valid max-heap as its left subtree
• A valid max-heap as its right subtree

We can create a valid max-heap rooted at v by "heapifying" (bubbling down) v to its correct position.

Why This Works:

When we bubble-down v:

1. If v is already largest among itself and children → heap is valid, we're done
2. If a child is larger → swap v with that child
3. The subtree where v lands is still a valid heap (by assumption), so we recurse

The recursion terminates because:

• Either v becomes larger than its children
• Or v reaches a leaf position (no children to compare)

At each step, the heap property is restored for the current level, and the subtrees remain valid heaps.

Visualizing Composition:

BEFORE (two valid heaps + arbitrary root):

        [3]         ← arbitrary value
       /   \
     [9]   [7]      ← roots of valid sub-heaps
     / \   / \
   [5][4][2][1]     ← these subtrees are valid heaps

AFTER (heapify root):

        [9]         ← 3 bubbled down
       /   \
     [5]   [7]      
     / \   / \
   [3][4][2][1]     ← 3 found its position

Trace of the heapify-down process:

1. Start: value 3 at root
2. Compare 3 with children 9, 7: max is 9 (left)
3. Swap 3 ↔ 9: now 3 is at left child position
4. Compare 3 with children 5, 4: max is 5 (left)
5. Swap 3 ↔ 5: now 3 is at left-left position
6. 3 has no children → done

Result: Valid max-heap!

Before we dive into the algorithm, let's solidify our understanding of how complete binary trees map to arrays. This mapping is what makes efficient heap operations possible.

The Level-Order Mapping:

A complete binary tree is stored in an array using level-order (breadth-first) indexing:

Tree Structure:            Array Representation:

        0                  Index: 0  1  2  3  4  5  6  7  8  9  10 11 12
       / \                 Value: ?  ?  ?  ?  ?  ?  ?  ?  ?  ?  ?  ?  ?
      /   \                       ↑  ↑  ↑  ↑ ← Level 0,1,2 (internal)
     1     2                            Level 3,... ← (leaves)
    / \   / \
   3   4 5   6         For n=13:
  / \ / \              - Internal nodes: indices 0 to 5 (⌊13/2⌋ - 1 = 5)
 7  8 9 10             - Leaf nodes: indices 6 to 12 (⌊13/2⌋ = 6 onward)

Index Formulas (0-based indexing):

Why These Formulas Matter for Heapify:

The build-heap algorithm will iterate through internal nodes from the last one to the root:

for i from ⌊n/2⌋ - 1 down to 0:
    heapify-down(i)

Let's verify the formula with examples:

Example 1: n = 10

⌊10/2⌋ - 1 = 5 - 1 = 4
Internal nodes: 0, 1, 2, 3, 4 (5 nodes)
Leaf nodes: 5, 6, 7, 8, 9 (5 nodes)

Tree:
          0
        /   \
       1      2
      / \    / \
     3   4  5   6
    / \ /
   7  8 9

Node 4's children: 2*4+1=9, 2*4+2=10 (10 is out of bounds)
  → Node 4 has one child (9) ✓
Node 5's children: 2*5+1=11 (out of bounds)
  → Node 5 is a leaf ✓

Example 2: n = 7 (complete, full tree)

⌊7/2⌋ - 1 = 3 - 1 = 2
Internal nodes: 0, 1, 2 (3 nodes)
Leaf nodes: 3, 4, 5, 6 (4 nodes)

This structural understanding—where leaves are, where internal nodes are, and how they relate—is the foundation for understanding why bottom-up heapify is so efficient.

We're now equipped to understand the two main approaches to heap construction. In the following pages, we'll analyze each in detail.

Approach 1: Naive (Top-Down) — O(n log n)

Insert elements one by one, each bubbling UP through potentially log(n) levels.

for i from 0 to n-1:
    // Treat A[0..i] as heap, add A[i+1]
    bubble-up(i+1)

Each element starts at the bottom and may travel to the root.

Approach 2: Optimal (Bottom-Up) — O(n)

Process internal nodes from bottom to top, each bubbling DOWN through limited levels.

for i from ⌊n/2⌋ - 1 down to 0:
    bubble-down(i)

Each element starts at its position and may travel to a leaf.

The Crucial Difference:

In the naive approach, elements near the bottom (which are numerous) may bubble all the way to the top (long distance).

In the optimal approach, elements near the bottom (which are numerous) only bubble down a short distance, while elements near the top (which are few) may bubble down a long distance.

This inversion of work distribution is what achieves O(n) instead of O(n log n).

We've established the foundation for understanding heap construction algorithms. Let's consolidate the key concepts before diving into the algorithms themselves.

What's Next:

In the following pages, we'll:

1. Page 2: Naive Approach — Analyze the repeated-insertion algorithm in detail, proving its O(n log n) complexity and understanding exactly why it's suboptimal.

2. Page 3: Bottom-Up Heapify — Present the optimal O(n) algorithm with full implementation, correctness proof, and step-by-step examples.

3. Page 4: Why Bottom-Up is Faster — Provide the mathematical analysis that proves O(n) complexity, with intuition for why the summation converges.

With the foundational understanding from this page, you're ready to appreciate the elegance and efficiency of the heapify algorithm.