Top-K with a Heap

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

Algorithm

@steps

Store the heap in an array.
Compare parent and child indexes instead of building explicit tree nodes.
Swap only when the heap order is violated.
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.

Java DSA Implementation

Basic.java

import java.util.*;
class Basic {
    static String listString(List<Integer> values) { return values.toString(); }
    static void heapInsert(List<Integer> heap, int value) {
        heap.add(value);
        int child = heap.size() - 1;
        while (child > 0) {
            int parent = (child - 1) / 2;
            if (heap.get(parent) <= heap.get(child)) break;
            Collections.swap(heap, parent, child);
            child = parent;
        }
    }
    static int heapPop(List<Integer> heap) {
        int smallest = heap.get(0);
        heap.set(0, heap.remove(heap.size() - 1));
        int parent = 0;
        while (true) {
            int left = parent * 2 + 1;
            int right = left + 1;
            if (left >= heap.size()) break;
            int child = left;
            if (right < heap.size() && heap.get(right) < heap.get(left)) child = right;
            if (heap.get(parent) <= heap.get(child)) break;
            Collections.swap(heap, parent, child);
            parent = child;
        }
        return smallest;
    }
    public static void main(String[] args) { List<Integer> heap = new ArrayList<>(); for (int value : new int[] {5, 1, 9, 3, 7, 2}) { heapInsert(heap, value); if (heap.size() > 3) heapPop(heap); } heap.sort(Collections.reverseOrder()); System.out.println(listString(heap)); }
}

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