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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main{

  public static void main(String[] args) {
    
    new Main().run();
  }

  private void run() {

    try(BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {

      int heapSize = Integer.parseInt(br.readLine());
      String[] keys = br.readLine().split(" ");
      int[] mHeap = new int[heapSize+1];
      mHeap[0] = heapSize;
      for(int i=1; i<=heapSize; i++)
        mHeap[i] = Integer.parseInt( keys[i-1] );

      buildMaxHeap(mHeap);

      StringBuilder buf = new StringBuilder();
      for(int i=1; i<mHeap.length; i++) 
        buf.append(" ").append(mHeap[i]);

      System.out.println(buf);
    } catch(IOException e) {
      System.out.println("IOException!");
    }
  }
  private void buildMaxHeap(int[] heap) {
    for(int i= heap[0]/2; i>0; i--)
      maxHeapify(heap, i);
  }
  private void maxHeapify(int[] heap, int node) {
    int t,left,right,largest;
    t = largest = 0;
    left  = nodeLeft(node);
    right = nodeRight(node);

    if( (left <= heap[0]) && (heap[left] > heap[node]) )
      largest = left;
    else
      largest = node;
    if( (right <= heap[0]) && (heap[right] > heap[largest]) )
      largest = right;

    if( largest != node ) {
      t = heap[node]; heap[node] = heap[largest]; heap[largest] = t;

      maxHeapify(heap, largest);
    }
  }
  private int nodeLeft(int node) {
    return node*2;
  }
  private int nodeRight(int node) {
    return node*2+1;
  }
}