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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.StringTokenizer;

public class Main {
	
	public static void maxHeapify(int[] A, int i, final int H){
		int l = i * 2 + 1;
		int r = i * 2 + 2;
		
		int largest = -1;
		
		if(l < H && A[l] > A[i]){
			largest = l;
		}else{
			largest = i;
		}
		
		if(r < H && A[r] > A[largest]){
			largest = r;
		}
		
		if(largest != i){
			int tmp = A[i];
			A[i] = A[largest];
			A[largest] = tmp;
			
			maxHeapify(A, largest, H);
		}
	}
	
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		
		final int n = sc.nextInt();
		int[] heap = new int[n];
		for(int i = 0; i < n; i++){
			heap[i] = sc.nextInt();
		}
		
		for(int i = n / 2; i >= 0; i--){
			maxHeapify(heap, i, n);
		}
		
		for(int i = 0; i < n; i++){
			System.out.print(" " + heap[i]);
		}
		System.out.println();
		
	}

	public static class Scanner {
		private BufferedReader br;
		private StringTokenizer tok;

		public Scanner(InputStream is) throws IOException {
			br = new BufferedReader(new InputStreamReader(is));
		}

		private void getLine() throws IOException {
			while (!hasNext()) {
				tok = new StringTokenizer(br.readLine());
			}
		}

		private boolean hasNext() {
			return tok != null && tok.hasMoreTokens();
		}

		public String next() throws IOException {
			getLine();
			return tok.nextToken();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public void close() throws IOException {
			br.close();
		}
	}

}