Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

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.Scanner;

class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		Node[] heap = new Node[n + 1];
		heap[0] = new Node();
		for (int i = 1; i < n + 1; i++) {
			heap[i] = new Node();
			heap[i].id = i;
			heap[i].key = sc.nextInt();
		}

		buildMaxHeap(heap, n);
		for(int i = 1;i < n+1;i++){
			System.out.print(" "+heap[i].key);
		}
		System.out.println();
	}

	static void maxHeapify(Node A[], int i, int n) {
		int l = 2 * i;
		int r = 2 * i + 1;
		int max = i;
		//System.out.println(l+" "+r);
		if (l <= n && A[l].key > A[max].key)max = l;
		if (r <= n && A[r].key > A[max].key)max = r;
		if (max != i) {
			int temp = A[i].key;
			A[i].key = A[max].key;
			A[max].key = temp;
			maxHeapify(A, max, n);
		}

	}

	static void buildMaxHeap(Node A[], int n){
		for(int i = n/2;i > 0;i--){
			maxHeapify(A, i, n);
		}
	}
}

class Node {
	int id;
	int key;
	int p;
	int l;
	int r;

	Node() {
		id = Integer.MIN_VALUE;
		key = Integer.MIN_VALUE;
		p = Integer.MIN_VALUE;
		l = Integer.MIN_VALUE;
		r = Integer.MIN_VALUE;
	}
}