find a heapified array when converting it to a sorted array, the total number of exchanges is maximal possible

Question

Inspired by this post, I googled the worst case of heapsort and found this question on cs.stackexchange.com, but the only answer didn't really answer the question, so I decided to dig it out myself. After hours of reasoning and coding, I've solved it. and I think this question belongs better in SO, so I post it up here.
The problem is to find a heapified array containing different numbers from 1 to n, such that when converting it to a sorted array, the total number of exchanges in all sifting operations is maximal possible.

Answer 1

Of course there is a brute force algorithm which calculates all possible of the heapified arrays and counts the number of exchanges for each one, and I have done that to verify the result of the solution below.

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• let's start from N=1: 1

• N=2: apparently, it's [2, 1]

• N=3: [3, x, 1].
  Since each sifting call will incur, at most, a number of swaps equal to the "height(which is equal to ⌊log(n)⌋" (from the bottom of the heap) of the node on which the sifting call is made, so we place 1 to the end of the array. apparently, x=2.

• N=4: [4, x, y, 1]
  After first extract-max, we need heapify [1, x, y]. If we sift it to the case when N=3, [3, 2, 1], since this sifting operation incurs the most swaps which is equal to the "height", plus the maximal number of exchanges when N=3, so that's the scenario of maximal number of exchanges when N=4. Thus, we do the "siftDown" version of heapify backwards to [3, 2, 1]: swap 1 with its parent until 1 is the root. So x=2, y=3

• N = n: [n,a,b,c,...,x,1]
  So, by induction, we do the same thing when N=n: after first extract-max, we sift down [1, a, b, c, ..., x] to the heapified array when N= n-1. So we do this backwards, get what we what.

Here is the code that outputs the heapified array which meets the requirements when you input N:

#include<stdio.h>

const int MAXN = 50001;

int  heap[MAXN];

int main()
{
    int n;
    int len,i,j;
    while(scanf("%d",&n)!=EOF)
    {
        heap[1]=1;
        len=1;
        for(i=2;i<=n;i++)
        {
            j=len;
            while(j>1)
            {
                heap[j]=heap[j/2];
                j/=2;
            }
            heap[1]=i;
            heap[++len]=1;
        }
        for(i=1;i<=n;i++)
        {
            if(i!=1) printf(" ");
            printf("%d",heap[i]);
        }
        printf("\n");
    }
    return 0;
}