08-heaps

Top-K with a Heap

Keep only the largest k values by maintaining a small min-heap.

Algorithm

Steps

  1. Store the heap in an array.
  2. Compare parent and child indexes instead of building explicit tree nodes.
  3. Swap only when the heap order is violated.
  4. Print the deterministic final heap state for replay comparison.
bounded heap For top-k largest values, a min-heap of size k keeps the current cutoff at the root.

Complexity

  • Time: O(n log k)
  • Space: O(k)

Dart DSA Implementation

basic.dart
Replay: real traced execution (multi-file project)
String listString(List<int> values) => "[${values.join(", ")}]";
void heapInsert(List<int> heap, int value) {
  heap.add(value);
  var child = heap.length - 1;
  while (child > 0) {
    final parent = (child - 1) ~/ 2;
    if (heap[parent] <= heap[child]) break;
    final tmp = heap[parent]; heap[parent] = heap[child]; heap[child] = tmp;
    child = parent;
  }
}
int heapPop(List<int> heap) {
  final smallest = heap[0];
  heap[0] = heap.removeLast();
  var parent = 0;
  while (true) {
    final left = parent * 2 + 1;
    final right = left + 1;
    if (left >= heap.length) break;
    var child = left;
    if (right < heap.length && heap[right] < heap[left]) child = right;
    if (heap[parent] <= heap[child]) break;
    final tmp = heap[parent]; heap[parent] = heap[child]; heap[child] = tmp;
    parent = child;
  }
  return smallest;
}
void main() { final heap = <int>[]; for (final value in [5, 1, 9, 3, 7, 2]) { heapInsert(heap, value); if (heap.length > 3) heapPop(heap); } heap.sort((a, b) => b.compareTo(a)); print(listString(heap)); }

  1. top-k heap ← [5]

    27}28void main() { final heap = <int>[]; for (final value in [5, 1, 9, 3, 7, 2]) { heapInsert(heap, value); if (heap.length > 3) heapPop(heap); } heap.sort((a, b) => b.compareTo(a)); print(listString(heap)); }
    values this step[5]top-k heap5value

  2. top-k heap ← [1, 5]

    27}28void main() { final heap = <int>[]; for (final value in [5, 1, 9, 3, 7, 2]) { heapInsert(heap, value); if (heap.length > 3) heapPop(heap); } heap.sort((a, b) => b.compareTo(a)); print(listString(heap)); }
    values this step[1, 5]top-k heap1value

  3. top-k heap ← [1, 5, 9]

    27}28void main() { final heap = <int>[]; for (final value in [5, 1, 9, 3, 7, 2]) { heapInsert(heap, value); if (heap.length > 3) heapPop(heap); } heap.sort((a, b) => b.compareTo(a)); print(listString(heap)); }
    values this step[1, 5, 9]top-k heap9value

  4. top-k heap ← [3, 5, 9]

    27}28void main() { final heap = <int>[]; for (final value in [5, 1, 9, 3, 7, 2]) { heapInsert(heap, value); if (heap.length > 3) heapPop(heap); } heap.sort((a, b) => b.compareTo(a)); print(listString(heap)); }
    values this step[3, 5, 9]top-k heap3value

  5. top-k heap ← [5, 7, 9]

    27}28void main() { final heap = <int>[]; for (final value in [5, 1, 9, 3, 7, 2]) { heapInsert(heap, value); if (heap.length > 3) heapPop(heap); } heap.sort((a, b) => b.compareTo(a)); print(listString(heap)); }
    values this step[5, 7, 9]top-k heap7value

  6. top-k heap ← [5, 7, 9]

    27}28void main() { final heap = <int>[]; for (final value in [5, 1, 9, 3, 7, 2]) { heapInsert(heap, value); if (heap.length > 3) heapPop(heap); } heap.sort((a, b) => b.compareTo(a)); print(listString(heap)); }
    values this step[5, 7, 9]top-k heap2value

  7. stdout ← [9, 7, 5]

    27}28void main() { final heap = <int>[]; for (final value in [5, 1, 9, 3, 7, 2]) { heapInsert(heap, value); if (heap.length > 3) heapPop(heap); } heap.sort((a, b) => b.compareTo(a)); print(listString(heap)); }
    values this step[9, 7, 5]stdout[5, 7, 9]top-k heap

Output

[9, 7, 5]