Problem:
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Maximum Heap

A binary heap which satisfies max-heap property is called max-heap. In a max-heap, for every node $i$ other than the root, $A[i] \leq A[parent(i)]$, that is, the value of a node is at most the value of its parent. The largest element in a max-heap is stored at the root, and the subtree rooted at a node contains values no larger than that contained at the node itself.

Here is an example of a max-heap.

Write a program which reads an array and constructs a max-heap from the array based on the following pseudo code.

$maxHeapify(A, i)$ move the value of $A[i]$ down to leaves to make a sub-tree of node $i$ a max-heap. Here, $H$ is the size of the heap.

1 maxHeapify(A, i)
2 l = left(i)
3 r = right(i)
4 // select the node which has the maximum value
5 if l ≤ H and A[l] > A[i]
6 largest = l
7 else
8 largest = i
9 if r ≤ H and A[r] > A[largest]
10 largest = r
11
12 if largest ≠ i　// value of children is larger than that of i
13 swap A[i] and A[largest]
14 maxHeapify(A, largest) // call recursively

The following procedure buildMaxHeap(A) makes $A$ a max-heap by performing maxHeapify in a bottom-up manner.

1 buildMaxHeap(A)
2 for i = H/2 downto 1
3 maxHeapify(A, i)

Input

In the first line, an integer $H$ is given. In the second line, $H$ integers which represent elements in the binary heap are given in order of node id (from $1$ to $H$).

Output

Print values of nodes in the max-heap in order of their id (from $1$ to $H$). Print a single space character before each value.

Constraint

$1 \leq H \leq 500,000$

$-2,000,000,000 \leq$ value of a node $\leq 2,000,000,000$

Sample Input 1

10
4 1 3 2 16 9 10 14 8 7

Sample Output 1

16 14 10 8 7 9 3 2 4 1

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in JAVA (judge verdict: Accepted)

import java.util.*;

class Main {
    class BinaryHeap {
        private long[] heap;
        private int size;

        public BinaryHeap(int capacity) {
            this.heap = new long[capacity];
            this.size = 0;
        }

        public void insert(long key) {
            this.heap[this.size] = key;
            this.size++;
        }

        public void dumpKeys() {
            for (int i = 0; i < this.size; i++) {
                System.out.printf(" %d", this.heap[i]);
            }
            System.out.print("\n");
        }

        public void buildMaxHeap() {
            for (int i = this.size / 2; i >= 0; i--) {
                this.maxHeapify(i);
            }
        }

        private void maxHeapify(int i) {
            int largest = i;

            int l = this.leftIndexOf(i);
            if (l < this.size && this.heap[l] > this.heap[largest]) {
                largest = l;
            }

            int r = this.rightIndexOf(i);
            if (r < this.size && this.heap[r] > this.heap[largest]) {
                largest = r;
            }

            if (largest != i) {
                long tmp = this.heap[i];
                this.heap[i] = this.heap[largest];
                this.heap[largest] = tmp;
                this.maxHeapify(largest);
            }
        }

        private int parentIndexOf(int i) {
            return ((i + 1) / 2) - 1;
        }

        public int leftIndexOf(int i) {
            return ((i + 1) * 2) - 1;
        }

        public int rightIndexOf(int i) {
            return (i + 1) * 2;
        }
    }
    public static void main(String[] args) {
        new Main().main();
    }

    public void main() {
        Scanner scanner = new Scanner(System.in);
        int n = Integer.parseInt(scanner.nextLine());
        String[] line = scanner.nextLine().split(" ", n);

        BinaryHeap heap = new BinaryHeap(n);
        for (int i = 0; i < n; i++) {
            long key = Long.parseLong(line[i]);
            heap.insert(key);
        }

        heap.buildMaxHeap();
        heap.dumpKeys();
    }
}