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: WA: Presentation Error)

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

public class Main {
    
    public static void main(String[] args) {
        Main main = new Main();
        main.solv();
    }
    
    int heapSize = 0;
    private void solv() {
        Scanner input = new Scanner(System.in);
        
        heapSize = input.nextInt();
        int[] heap = new int[heapSize+1];
        for (int i = 1; i < heapSize+1; i++) {
            heap[i] = input.nextInt();
        }
        buildMaxHeap(heap);
        StringBuilder sb = new StringBuilder();
        for (int i = 1; i < heapSize+1; i++) {
            sb.append(heap[i]).append(' ');
        }
        System.out.println(sb.toString().trim());
    }
    int parent(int i) {
        return i / 2;
    }
    int left(int i) {
        return i * 2;
    }
    int right(int i) {
        return i * 2 + 1;
    }
    void buildMaxHeap(int[] A) {
        for (int i = heapSize/2; i > 0; i--) {
            maxHeapify(A, i);
        }
    }
    void maxHeapify(int[] A, int i) {
        int l = left(i);
        int r = right(i);
        int largest;
        if (l <= heapSize && A[l] > A[i]) {
            largest = l;
        } else {
            largest = i;
        }
        if (r <= heapSize && A[r] > A[largest]) largest = r;
        if (largest != i) {
            swap(A, i, largest);
            maxHeapify(A, largest);
        }
    }
    void swap(int[] A, int x, int y) {
        int temp = A[x];
        A[x] = A[y];
        A[y] = temp;
    }
    
    private static class Scanner {
        BufferedReader r;
        StreamTokenizer st;
        
        public Scanner(InputStream in) {
            r = new BufferedReader(new InputStreamReader(in));
            st = new StreamTokenizer(r);
        }
        
        public String next() {
            nextToken();
            return st.sval;
        }
        
        public Integer nextInt() {
            nextToken();
            return (int)st.nval;
        }
        
        private int nextToken() {
            int token = st.TT_EOF;
            try {
                token = st.nextToken();
            } catch (IOException e) {
            }
            return token;
        }
    }
}