Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

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 Python3 (judge verdict: WA: Presentation Error)

import math

def parent(i):
    return math.floor(i / 2)

def left(i):
    return 2 * i

def right(i):
    return 2 * i + 1

def get_largest(A, n, i):
    l = left(i)
    r = right(i)
    largest = None
    if l <= n and A[i] < A[l]:
        largest = l
    else:
        largest = i
    if r <= n and A[largest] < A[r]:
        largest = r
    return largest


def max_heapify(A, n, i):
    largest = get_largest(A, n, i)
    x = i
    while largest is not x:
        tmp = A[x]
        A[x] = A[largest]
        A[largest] = tmp
        x = largest
        largest = get_largest(A, n, x)

def build_max_heap(A, n):
    for i in reversed(range(1, math.floor(n/2) + 1)):
        max_heapify(A, n, i)

n = int(input())
heap = [None] + list(map(int, input().split()))

build_max_heap(heap, n)

print(" ", *heap[1:])