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.Scanner;


public class Main{

	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int cnt = Integer.parseInt(s.nextLine());
		int[] maxHeap = new int[cnt];
		String[] tmp = s.nextLine().split(" ");
		for(int i=0 ; i<cnt ; i++) {
			maxHeap[i] = Integer.parseInt(tmp[i]);
		}
		buildMaxHeap(maxHeap);
		for(int i=0 ; i<maxHeap.length ; i++) {
			System.out.print(" " + maxHeap[i]);
		}
		System.out.println("");
	}

	public static void maxHeapify(int[] maxHeap, int index) {
		int leftChild = index*2;
		int rightChild = (index*2)+1;
		int max = maxHeap[index-1];
		if(maxHeap.length>=leftChild && maxHeap[leftChild-1]>maxHeap[index-1]) {
			max = maxHeap[leftChild-1];
		}
		if(maxHeap.length>=rightChild && maxHeap[rightChild-1]>max) {
			max = maxHeap[rightChild-1];
		}
		if(max!=maxHeap[index-1]) {
			if(max==maxHeap[leftChild-1]) {
				int tmp = maxHeap[leftChild-1];
				maxHeap[leftChild-1] = maxHeap[index-1];
				maxHeap[index-1] = tmp;
				maxHeapify(maxHeap, leftChild);
			}else if(max==maxHeap[rightChild-1]) {
				int tmp = maxHeap[rightChild-1];
				maxHeap[rightChild-1] = maxHeap[index-1];
				maxHeap[index-1] = tmp;
				maxHeapify(maxHeap, rightChild);
			}
		}
	}

	public static void buildMaxHeap(int[] maxHeap) {
		for(int i=(maxHeap.length/2) ; i>0 ; i--) {
			maxHeapify(maxHeap, i);
		}
	}
}