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;
import java.util.*;
import java.util.Arrays;

import java.io.*;

public class Main {
    static int MAX = 4000000;
    
    static int[] A = new int[MAX+10];
    static int F;
    static void maxHeap(int i) {
        int l, r, huge;
        r = 2 * i + 1;
        l = 2 * i;
        
        if(l <= F && A[l] > A[i]) huge = l;
        else huge = i;
        if(r <= F && A[r] > A[huge]) huge = r;

        if(huge != i) {
            int temp = A[i];
            A[i] = A[huge];
            A[huge] = temp;
            maxHeap(huge);
        }
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        count(sc);
    }
 public static void count(Scanner s){
  F = s.nextInt();
        for(int i = 1; i <= F; ++i) A[i] = s.nextInt();

        for(int i = F / 2; i >= 1; i--) {
        maxHeap(i);
}
        for(int K = 1; K <= F; K++) {
            System.out.print(" " + A[K]);
        }
        System.out.println();
    }
 
 
 }