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;

class Main
{
    static int N;
    static int[]A = new int[10000000];

    public static void maxHeapify(int i)
    {
	int l = 2 * i;
	int r = 2 * i + 1;
	int largest;

	if(l <= N && A[l] > A[i])
	    largest = l;
	else 
	    largest = i;

	if(r <= N && A[r] > A[largest])
	    largest = r;

	if(largest != i)
	    {
		int b = A[i];
		A[i] = A[largest];
		A[largest] = b;
		maxHeapify(largest);
	    }
    }

    public static void main(String[] args)
    {
	Scanner In = new Scanner(System.in);
	    

	N = In.nextInt();

	for(int i = 1; i <= N; i++)
	    A[i] = In.nextInt();
	
	for(int i = N / 2; i >= 1; i--)
	    maxHeapify(i);
	
	for(int i = 1; i <= N; i++)
	    System.out.print(" " + A[i]);
	
	System.out.println();
    }
}