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: Runtime Error)

import java.util.*;
import java.io.*;

class Main{
    static int largest;
    static int l,r,H;
    static int[] A;
    public static void main(String args[]){
	int i,j;
	A = new int[500000];
	try{
	    BufferedReader buffer = new BufferedReader(new InputStreamReader(System.in));
	    String str = buffer.readLine();
	    H = Integer.parseInt(str);
	    StringTokenizer st = new StringTokenizer(buffer.readLine());
	    for(i=1;i<=H;i++){
		A[i] = Integer.parseInt(st.nextToken());
	    }
	}catch(Exception e){
	    System.out.println("error");
	}
	buildMaxHeap(A);
	StringBuffer sb = new StringBuffer();
	for(i=1;i<=H;i++){
	    sb.append(" "+A[i]);
	}
	System.out.println(sb);	
    }

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

    static int left(int x){
	return 2*x;
    }
    static int right(int x){
	return 2*x+1;
    }
    static void maxHeapify(int A[],int i){
	int temp;
	l = left(i);
	r = right(i);
	if(l <= H && A[l] > A[i]){
	    largest = l;
	}
	else{largest = i;}

	if(r <= H && A[r] > A[largest]){
	    largest = r;
	}
	if(largest != i){
	    temp = A[i];
	    A[i] = A[largest];
	    A[largest] = temp;
	    maxHeapify(A,largest);
	}
    }
}