Problem:
Score : 400 points

Problem Statement

You are given N integers; the i-th of them is A_i.
Find the maximum possible sum of the absolute differences between the adjacent elements after arranging these integers in a row in any order you like.

Constraints

2 \leq N \leq 10^5

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

Print the maximum possible sum of the absolute differences between the adjacent elements after arranging the given integers in a row in any order you like.

Sample Input 1

5
6
8
1
2
3

Sample Output 1

21

When the integers are arranged as 3,8,1,6,2, the sum of the absolute differences between the adjacent elements is |3 - 8| + |8 - 1| + |1 - 6| + |6 - 2| = 21. This is the maximum possible sum.

Sample Input 2

6
3
1
4
1
5
9

Sample Output 2

25

Sample Input 3

3
5
5
1

Sample Output 3

8

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5+110;
const long long mod = 998244353;
typedef long long ll;

int n;
ll a[maxn];
ll _abs(ll x)
{
    if(x<0)return -x;
    return x;
}

int main()
{
    scanf("%d", &n);
    for(int i=1;i<=n;i++)
    {
        scanf("%lld", &a[i]);
    }
    sort(a+1, a+n+1);
    if(n==2)
    {
        printf("%lld\n" , a[2]-a[1]);
        return 0;
    }
    int l=2, r=n-2;
    ll sum=a[n]-a[1]+a[n-1]-a[1];
    ll left=a[n-1];
    ll right=a[n];
    int cnt=0;
    while(l<=r)
    {
        if(cnt==1)
        {
            sum += _abs(left-a[l]);
            left=a[l];
            ++cnt;
            ++l;
        }
        else if(cnt==0)
        {
            sum += _abs(right-a[l]);
            right = a[l];
            ++l;
            ++cnt;
        }
        else if(cnt==3)
        {
            sum += _abs(left-a[r]);
            left=a[r];
            --r;
            ++cnt;
        }
        else if(cnt==2)
        {
            sum += _abs(right-a[r]);
            right=a[r];
            --r;
            ++cnt;
        }
        cnt%=4;
    }
    printf("%lld\n", sum);
    return 0;

}