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. @end @complexity
  • Time: O(n log k)
  • Space: O(k) @end
bounded heap For top-k largest values, a min-heap of size k keeps the current cutoff at the root.

Dart DSA Implementation

basic.dart
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)); }

@end @output [9, 7, 5] @end