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.*;
import static java.lang.System.*;
import static java.lang.Math.*;

public class Main {
  static int h;

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

    h = Integer.parseInt(sc.nextLine()) + 1;
    int[] a = new int[h];

    int[] input = Arrays.stream(sc.nextLine().split(" ")).mapToInt(Integer::parseInt).toArray();
    for (int i = 1; i < h; i++) {
        a[i] = input[i-1];
    }

    buildMaxHeap(a);
    for (int i = 1; i < h; i++) {
      System.out.print(" " + a[i]);
    }

    System.out.println();
  }

  static void buildMaxHeap (int[] a) {
    for (int i = h/2; i >= 1; i--) {
      maxHeapify(a, i);
    }
  }

  static void maxHeapify (int[] a, int i) {
    int leftId = i * 2;
    int rightId = i * 2 + 1;
    int largest = 0;

    if (leftId < h && a[leftId] > a[i]) {
	    largest = leftId;
    } else {
      largest = i;
    }

    if (rightId < h && a[rightId] > a[largest]) {
      largest = rightId;
    }

    if (largest != i) {
      int tmp = a[i];
      a[i] = a[largest];
      a[largest] = tmp;
      maxHeapify(a, largest);
    }
  }
}