Heap Sort

Sorting O(n log n) O(1)

Builds a max-heap, then repeatedly swaps the root to the end and re-heapifies the shrinking prefix.

Idle

Starting Heap Sort on 8 elements.

InvariantEvery subtree below the current index already satisfies the heap property, so sifting one node down is enough.

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comparisons0swaps0size8i0largest0

compareswapshiftselectsorted

heap-sort.c

void heapify(int arr[], int n, int i) {
    int largest = i;
    int l = 2 * i + 1, r = 2 * i + 2;
    if (l < n && arr[l] > arr[largest]) largest = l;
    if (r < n && arr[r] > arr[largest]) largest = r;
    if (largest != i) {
        int tmp = arr[i];
        arr[i] = arr[largest];
        arr[largest] = tmp;
        heapify(arr, n, largest);
    }
}
 
void heapSort(int arr[], int n) {
    for (int i = n / 2 - 1; i >= 0; i--)
        heapify(arr, n, i);
    for (int i = n - 1; i > 0; i--) {
        int tmp = arr[0];
        arr[0] = arr[i];
        arr[i] = tmp;
        heapify(arr, i, 0);
    }
}

How Heap Sort works

Build a max-heap over the array, then repeatedly swap the root (maximum) with the last element and sift the new root down over the shrinking prefix.

Best case

Always O(n log n); heap build itself is only O(n).

Worst case

Still O(n log n) — no data-dependent blowup, and O(1) extra space.

When to use it

When you need guaranteed O(n log n) in place (real-time / adversarial inputs) and stability doesn't matter.