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.*;
import java.io.*;
public class Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int count = sc.nextInt();
        int[] arr = new int[count+1];
        for(int i=1;i<=count;i++){
            arr[i] = sc.nextInt();
        }
        for(int i=count/2;i>0;i--){
            MaxHeap(arr,i);
        }

        for(int i=1;i<=count;i++){
            System.out.print(" "+arr[i]);
        }
        System.out.println();

    }
/*
     3
  11     6
9   4
 */
    public static void MaxHeap(int[] arr,int i){//3 4 6 9 11
       int left = 2*i;
       int right = 2*i+1;
       int largest;
       if(left<=arr.length-1&&arr[left]>arr[i]){
           largest = left;
       }else{
           largest = i;
       }
       if(right<=arr.length-1&&arr[right]>arr[largest]){
           largest = right;
       }
       if(largest!=i){
           swap(arr,i,largest);
           MaxHeap(arr,largest);
       }

    }

    public static void swap(int a[], int i, int largest){
        int temp = a[i];
        a[i] = a[largest];
        a[largest] = temp;
    }


}