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: Accepted)

#include <bits/stdc++.h>
#include <math.h>
#include <time.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
const ll mod = 1000000007;
const int INF = 1001001001;
const ll LINF = 1001001001001001001;

int main(){
    ll n ; cin >> n;
    vector<ll> a(n);
    rep(i,n) cin >> a.at(i);
    sort(a.begin(),a.end());

    ll x = a.at(n-1);
    ll ans1 = 0;
    ll first=x,second=x,goukei=2*x;
    for(int i=0;i<(n-1)/2;i++){
        if(i%2){
            ll m = 1+(i/2);
            first = a.at(n-2*m);
            second = a.at(n-2*m-1);
            ans1 += abs(first+second-goukei); 
        }else{
            ll m = i/2;
            first = a.at(2*m);
            second = a.at(2*m+1);
            ans1 += abs(first+second-goukei);            
        }
        goukei = first+second;
    }
    if(n%4==2){
        ll y = a.at(n/2 - 1);
        ans1 += max(abs(first-y),abs(second-y));
    }else if(n%4==0){
        ll y = a.at(n/2);
        ans1 += max(abs(first-y),abs(second-y));
    }

    x = a.at(0);
    ll ans2 = 0;
    first=x,second=x,goukei=2*x;
    for(int i=0;i<(n-1)/2;i++){
        if(i%2==0){
            ll m = 1+(i/2);
            first = a.at(n-2*m+1);
            second = a.at(n-2*m);
            ans2 += abs(first+second-goukei); 
        }else{
            ll m = i/2;
            first = a.at(2*m+1);
            second = a.at(2*m+2);
            ans2 += abs(first+second-goukei);            
        }
        goukei = first+second;
    }
    if(n%4==0){
        ll y = a.at(n/2 - 1);
        ans2 += max(abs(first-y),abs(second-y));
    } else if(n%4==2){
        ll y = a.at(n/2);
        ans2 += max(abs(first-y),abs(second-y));
    }
  
    //cout << ans1 << ans2 << endl;
    cout << max(ans1,ans2) << endl;
}