When faced with the problem of building a heap from an unsorted array, the most natural solution draws on what we already know: heap insertion. We've mastered inserting a single element into a heap in O(log n) time, so why not simply start with an empty heap and insert each element?

This approach is intuitive, easy to implement, and correct—it absolutely produces a valid heap. However, it's not optimal. Understanding precisely why it's suboptimal is crucial, because this understanding illuminates why the optimal bottom-up algorithm works and why it's faster.

In this page, we'll dissect the naive approach completely: its algorithm, its correctness, its exact complexity, and the distribution of work that makes it inefficient. By the end, you'll have a crystal-clear understanding of the O(n log n) baseline that the optimal algorithm improves upon.

The naive heap construction algorithm mimics what you would do if you were building a heap incrementally: start empty, add one element at a time.

Algorithm: BUILD-HEAP-NAIVE (for max-heap)

BUILD-HEAP-NAIVE(A):
    n = length(A)
    
    // Conceptually, A[0] alone is already a valid heap of size 1
    // We "insert" elements A[1], A[2], ..., A[n-1] into this growing heap
    
    for i from 1 to n - 1:
        BUBBLE-UP(A, i)
    
    return A  // A is now a valid heap

BUBBLE-UP(A, index):
    while index > 0:
        parent = (index - 1) / 2  // integer division
        if A[index] > A[parent]:
            swap A[index] and A[parent]
            index = parent
        else:
            break

Key Observation:

This algorithm works in-place. We don't create a new array; we reinterpret the input array as a growing heap:

• After processing index i, the subarray A[0..i] is a valid max-heap
• Before processing, A[0..0] (just the first element) is trivially a valid heap
• The loop "extends" this heap to include one more element each iteration

In-Place Visualization:

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

Step 0: Heap is A[0..0] = [3]
        ✓ A single element is a valid heap

Step 1: Insert A[1]=9 into heap A[0..0]
        Compare 9 with parent 3: 9 > 3, swap
        Heap is now [9, 3]
        Array: [9, 3, 2, 1, 4, 5]

Step 2: Insert A[2]=2 into heap A[0..1]
        Compare 2 with parent 9: 2 < 9, done
        Heap is now [9, 3, 2]
        Array: [9, 3, 2, 1, 4, 5]

Step 3: Insert A[3]=1 into heap A[0..2]
        Compare 1 with parent 3: 1 < 3, done
        Heap is now [9, 3, 2, 1]
        Array: [9, 3, 2, 1, 4, 5]

Step 4: Insert A[4]=4 into heap A[0..3]
        Compare 4 with parent 3: 4 > 3, swap
        Heap is now [9, 4, 2, 1, 3]
        Compare 4 with parent 9: 4 < 9, done
        Array: [9, 4, 2, 1, 3, 5]

Step 5: Insert A[5]=5 into heap A[0..4]
        Compare 5 with parent 2: 5 > 2, swap
        Heap is now [9, 4, 5, 1, 3, 2]
        Compare 5 with parent 9: 5 < 9, done
        Array: [9, 4, 5, 1, 3, 2]

Final Heap: [9, 4, 5, 1, 3, 2] ✓

To rigorously prove that the naive algorithm produces a valid heap, we employ loop invariant analysis—the gold standard for algorithm correctness proofs.

Loop Invariant:

At the start of each iteration of the for loop, A[0..i-1] is a valid max-heap containing the first i elements of the original array (in some order).

Initialization (before first iteration, i = 1):

A[0..0] contains a single element. A single-element array trivially satisfies the max-heap property—there are no parent-child pairs to violate it.

✓ Invariant holds initially.

Maintenance (if invariant holds at start of iteration i, it holds at start of iteration i+1):

Assume A[0..i-1] is a valid max-heap before iteration i.

During iteration i, we call BUBBLE-UP(A, i). This operation:

1. Takes element A[i] (which is outside the current heap)
2. Compares it with its parent and swaps if necessary
3. Continues until A[i] reaches a position where it doesn't violate the heap property

We proved in the previous module that BUBBLE-UP correctly restores the heap property. After BUBBLE-UP completes, A[0..i] is a valid max-heap.

✓ Invariant maintained.

Termination (after loop completes, i = n):

When the loop terminates, i = n, so by the invariant, A[0..n-1] is a valid max-heap. This is the entire array.

✓ Algorithm is correct.

Now we establish that the naive algorithm is O(n log n). This analysis is critical because it reveals the exact nature of the inefficiency.

Per-Insertion Cost:

When we insert the element at index i (for i ≥ 1):

• The heap currently has i elements (indices 0 to i-1)
• The height of this heap is ⌊log₂(i)⌋
• In the worst case, BUBBLE-UP swaps through all levels: ⌊log₂(i)⌋ swaps

Total Cost:

The total work is the sum of worst-case costs for each insertion:

T(n) = Σ (i=1 to n-1) ⌊log₂(i)⌋

Let's compute this sum.

Lower Bound:

T(n) ≥ Σ (i=1 to n-1) (log₂(i) - 1)
     = Σ (i=1 to n-1) log₂(i) - (n-1)
     = log₂((n-1)!) - (n-1)

Upper Bound:

T(n) ≤ Σ (i=1 to n-1) log₂(i)
     = log₂(1) + log₂(2) + ... + log₂(n-1)
     = log₂(1 × 2 × 3 × ... × (n-1))
     = log₂((n-1)!)

Using Stirling's Approximation:

Stirling's approximation states that for large n:

log₂(n!) ≈ n log₂(n) - n log₂(e) + O(log n)
         = n log₂(n) - 1.443n + O(log n)

Applying this:

log₂((n-1)!) ≈ (n-1) log₂(n-1) - 1.443(n-1)
             = Θ(n log n)

Therefore:

T(n) = Θ(n log n)

Alternative Direct Analysis:

We can also analyze by counting nodes at each level and their potential work:

Level 0: 1 node,  can bubble up 0 levels
Level 1: 2 nodes, can bubble up 1 level each → 2 × 1 = 2
Level 2: 4 nodes, can bubble up 2 levels each → 4 × 2 = 8
Level 3: 8 nodes, can bubble up 3 levels each → 8 × 3 = 24
...
Level k: 2^k nodes, can bubble up k levels each → k × 2^k

If the tree has height h = log₂(n), total worst-case work:

T(n) = Σ (k=0 to h) k × 2^k

This sum evaluates to:

Σ (k=0 to h) k × 2^k = 2 + (h-1) × 2^(h+1) ≈ h × 2^h = log(n) × n

Confirming T(n) = O(n log n).

The key to understanding why the naive approach is inefficient lies in where the work is concentrated. Let's visualize this.

For a heap of n = 15 elements (height 3):

Level 0 (root):     1 node  × max 0 swaps = 0 work
Level 1:            2 nodes × max 1 swap  = 2 work
Level 2:            4 nodes × max 2 swaps = 8 work
Level 3 (leaves):   8 nodes × max 3 swaps = 24 work
                                           -------
Total worst-case work:                      34 operations

Observation: Level 3 (the leaves) contributes 24 out of 34 operations—over 70% of the work!

This pattern holds generally. The deepest level contains approximately half of all nodes, and each of those nodes can bubble all the way to the root.

The Fundamental Inefficiency:

The problem is structural. In the naive approach:

1. Most nodes are at the bottom — Approximately n/2 nodes are leaves
2. Bottom nodes travel the farthest — They can bubble up through log(n) levels
3. Work = nodes × distance — (n/2) × log(n) = Ω(n log n)

This is the worst possible distribution! We're making the most numerous nodes do the most work.

Mathematical Formulation:

Total Work = Σ (level k) (nodes at k) × (distance from k to root)
           = Σ (k=0 to h) 2^k × k
           = Θ(n log n)

If we could somehow invert this—make the numerous nodes do little work and the few nodes do more work—we could potentially achieve better complexity. This is exactly what the bottom-up heapify algorithm does.

Let's trace through the naive algorithm on a larger example to fully internalize the process and see the work distribution in action.

Input Array: A = [4, 1, 3, 2, 16, 9, 10, 14, 8, 7]

Expected: A valid max-heap containing these 10 elements

Step-by-Step Trace:

Final Array: [16, 14, 10, 4, 7, 3, 9, 1, 8, 2]

Verification (Max-Heap Property):

             16
           /    \
         14      10
        /  \    /  \
       4    7  3    9
      / \  /
     1  8 2

Parent-Child Checks:
- 16 ≥ 14, 16 ≥ 10 ✓
- 14 ≥ 4, 14 ≥ 7 ✓
- 10 ≥ 3, 10 ≥ 9 ✓
- 4 ≥ 1, 4 ≥ 8 ✗ WAIT, 4 < 8!

Oops! Let me re-trace more carefully...

Actually, let me redo step 7:

i=7: Insert 14 at index 7. Parent index = (7-1)/2 = 3. A[3]=2. 14 > 2, swap. Now 14 is at index 3, 2 is at index 7. Array: [16,4,10,14,2,3,9,2]. Wait, there's an error. Let me start fresh with careful tracking.

Corrected Trace (starting from i=6):

After i=6: [16, 4, 10, 1, 2, 3, 9]

i=7: Insert 14 at index 7.

• Parent index = (7-1)/2 = 3, A[3] = 1
• 14 > 1, swap → [16, 4, 10, 14, 2, 3, 9, 1]
• New index = 3, Parent = (3-1)/2 = 1, A[1] = 4
• 14 > 4, swap → [16, 14, 10, 4, 2, 3, 9, 1]
• New index = 1, Parent = (1-1)/2 = 0, A[0] = 16
• 14 < 16, stop
• Result: [16, 14, 10, 4, 2, 3, 9, 1] (2 swaps)

i=8: Insert 8 at index 8.

• Parent index = (8-1)/2 = 3, A[3] = 4
• 8 > 4, swap → [16, 14, 10, 8, 2, 3, 9, 1, 4]
• New index = 3, Parent = 1, A[1] = 14
• 8 < 14, stop
• Result: [16, 14, 10, 8, 2, 3, 9, 1, 4] (1 swap)

i=9: Insert 7 at index 9.

• Parent index = (9-1)/2 = 4, A[4] = 2
• 7 > 2, swap → [16, 14, 10, 8, 7, 3, 9, 1, 4, 2]
• New index = 4, Parent = 1, A[1] = 14
• 7 < 14, stop
• Result: [16, 14, 10, 8, 7, 3, 9, 1, 4, 2] (1 swap)

Total Swaps: 0+0+0+1+2+1+1+2+1+1 = 9 swaps

For n=10 elements, log₂(10. ≈ 3.3, so the naive approach's theoretical worst case would be ~10 × 3 = 30 swaps. Our actual count of 9 is well under because not every element needed to bubble all the way up.

Here are production-quality implementations of the naive heap construction algorithm in several languages.

Python Implementation:

def build_heap_naive(arr: list) -> list:
    """
    Build a max-heap from an array using repeated insertions.
    Time: O(n log n), Space: O(1) auxiliary
    """
    n = len(arr)
    
    # Process each element starting from index 1
    # (index 0 is trivially a valid heap of size 1)
    for i in range(1, n):
        _bubble_up(arr, i)
    
    return arr

def _bubble_up(arr: list, index: int) -> None:
    """Bubble element at index up to restore heap property."""
    while index > 0:
        parent = (index - 1) // 2
        
        if arr[index] > arr[parent]:
            arr[index], arr[parent] = arr[parent], arr[index]
            index = parent
        else:
            break

# Example usage
arr = [4, 1, 3, 2, 16, 9, 10, 14, 8, 7]
build_heap_naive(arr)
print(arr)  # [16, 14, 10, 8, 7, 3, 9, 1, 4, 2]

TypeScript Implementation:

function buildHeapNaive<T>(arr: T[], compare: (a: T, b: T) => number = (a, b) => (a > b ? 1 : -1)): T[] {
    const n = arr.length;
    
    for (let i = 1; i < n; i++) {
        bubbleUp(arr, i, compare);
    }
    
    return arr;
}

function bubbleUp<T>(arr: T[], index: number, compare: (a: T, b: T) => number): void {
    while (index > 0) {
        const parent = Math.floor((index - 1) / 2);
        
        if (compare(arr[index], arr[parent]) > 0) {
            [arr[index], arr[parent]] = [arr[parent], arr[index]];
            index = parent;
        } else {
            break;
        }
    }
}

// Example usage
const arr = [4, 1, 3, 2, 16, 9, 10, 14, 8, 7];
buildHeapNaive(arr);
console.log(arr);  // [16, 14, 10, 8, 7, 3, 9, 1, 4, 2]

Java Implementation:

public class NaiveHeapBuilder {
    public static void buildMaxHeap(int[] arr) {
        int n = arr.length;
        
        for (int i = 1; i < n; i++) {
            bubbleUp(arr, i);
        }
    }
    
    private static void bubbleUp(int[] arr, int index) {
        while (index > 0) {
            int parent = (index - 1) / 2;
            
            if (arr[index] > arr[parent]) {
                // Swap
                int temp = arr[index];
                arr[index] = arr[parent];
                arr[parent] = temp;
                index = parent;
            } else {
                break;
            }
        }
    }
}

The naive algorithm's space complexity is excellent—it's one of its few advantages.

Auxiliary Space: O(1)

The algorithm uses only:

• A few index variables (i, index, parent)
• A temporary variable for swapping (in some implementations)

No additional arrays, no recursion stack (the while loop is iterative), no extra data structures.

In-Place Property:

The algorithm modifies the input array directly. After execution:

• The original array contains a valid heap
• No memory allocation occurred beyond function call overhead
• The original values are preserved (just reordered)

Comparison with "Create New Heap" Approach:

An alternative naive approach creates a new heap and copies elements:

def build_heap_copy(arr):
    heap = []
    for x in arr:
        heap.append(x)
        bubble_up(heap, len(heap) - 1)
    return heap

This uses O(n) extra space for the new array. Our in-place version avoids this.

Despite being suboptimal, the naive approach is sometimes perfectly reasonable. Understanding when can save you from premature optimization.

Use Naive When:

Avoid Naive When:

• Large n — For n > 100,000, the difference becomes seconds vs. milliseconds
• Repeated builds — If you rebuild heaps frequently, the overhead accumulates
• Performance-critical paths — In hot loops, tight latency requirements, or competitive programming
• HeapSort implementation — The sort's overall efficiency depends on O(n) build
• Real-time systems — Predictable, minimal initialization time is essential

The Practical Threshold:

As a rough guideline:

• n < 10,000: Naive is fine, difference is ~10× (milliseconds either way)
• n ~ 100,000: Optimal becomes noticeable (~17× difference)
• n > 1,000,000: Optimal is essential (~20× difference, seconds vs. tenths of seconds)

We've thoroughly examined the naive heap construction algorithm. Let's consolidate the key insights.

The Core Insight for Improvement:

The naive approach is inefficient because:

Work = Σ (level k) [nodes at k] × [distance to root]
     = Σ (k=0 to h) 2^k × k
     = O(n log n)

If we could compute:

Work = Σ (level k) [nodes at k] × [distance to leaves]
     = Σ (k=0 to h) 2^k × (h - k)

This would give us a much smaller total—actually O(n). This is the bottom-up heapify approach, which we'll cover in the next page.

What's Next:

Now that we understand why the naive approach is O(n log n), we're ready to learn the optimal algorithm. The next page presents bottom-up heapify: the elegant O(n) algorithm that builds a heap by processing internal nodes from bottom to top, exploiting the favorable work distribution we identified.