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 <algorithm>
#include <iomanip>
#include <string>
#include <functional>
#include <utility>
#include <queue>
#include <vector>
#include <string>
#include <stack>
#define rep(i,n) for(ll i=0;i<n;i++)
using namespace std;
typedef long long int ll;
const ll  MOD = 1000000007;
typedef pair<ll,ll> P;
const ll INF = 100100100100100100;

ll N;
ll a[200200];
int main(){
  cin >> N;
  rep(i,N){
    cin >> a[i];
  }
  sort(a,a+N);

  ll ans = 0;
  ll left;
  ll right;
  rep(i,N){
    if(i==0){
      left = a[i];
      right = a[i];
    }
    else {
      ll elem = (i % 2 == 0 ? a[i/2]:a[N - i/2 - 1]);
      ll L = abs(left - elem);
      ll R = abs(right - elem);
      if (L == R && i != 1) {
        if (i%2 == 0){
          if(right > left) {
            ans += R;
            right = elem;
          }
          else{
            ans += L;
            left = elem;
          }
        }
      }
      else if(L > R){
        ans += L;
        left = elem;
      }
      else {
        ans += R;
        right = elem;
      }
    }
  }
  ll ans2 = 0;
  rep(i,N){
    if(i==0){
      left = a[N-1];
      right = a[N-1];
    }
    else {
      ll elem = (i % 2 == 1 ? a[i/2]:a[N - i/2 - 1]);
      ll L = abs(left - elem);
      ll R = abs(right - elem);
      if (L == R && i != 1) {
        if (i%2 == 0){
          if(right > left) {
            ans2 += L;
            left = elem;
          }
          else{
            ans2 += R;
            right = elem;
          }
        }
      }
      else if(L > R){
        ans2 += L;
        left = elem;
      }
      else {
        ans2 += R;
        right = elem;
      }
    }
  }

  cout << max(ans,ans2) << endl;
}