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<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
int num[100005];
int s[100005];
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>num[i];
    }
    sort(num,num+n);
    int mid=n/2;
    int j=n-1;
    int i=0;
    int sum=0;
    for(int t=0;t<n;t++)
    {
        if(t%2==0) s[t]=num[j--];
        else s[t]=num[i++];
//        cout<<s[t];
    }
//    cout<<endl;
    for(int i=1;i<n-1;i++)
    {
        sum+=abs(s[i]-s[i-1]);
    }
    int a=s[0]-s[n-1];
    int b=abs(s[n-1]-s[n-2]);
    if(a>b) sum+=a;
    else sum+=b;

    cout<<sum<<endl;
}