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<cmath>
#include<stack>
#include<queue>
#include<vector>

using namespace std;

int main() {
  int N;
  cin >> N;

  vector<long> A(N);
  for (int i=0; i<N; i++) cin >> A[i];

  sort(A.begin(), A.end());
  long res = 0;

  int left = 1;
  int right = A.size() - 1;
  int left_v = A[0];
  int right_v = A[0];

  bool left_f = false;
  int counter = 0;

  //cout << "-------" << endl;
  
  while (left<=right) {
    if (counter == 2) {
      counter = 0;
      if (left_f) left_f = false;
      else left_f = true;
    }

    if (counter == 0) {
      if (left_f) {
        res += abs(left_v - A[left]);
        left_v = A[left];
        left++;
      } else {
        res += abs(left_v - A[right]);
        left_v = A[right];
        right--;
      }
    }

    if (counter == 1) {
      if (left_f) {
        res += abs(right_v - A[left]);
        right_v = A[left];
        left++;
      } else {
        res += abs(right_v - A[right]);
        right_v = A[right];
        right--;
      }
    }

    // cout << "left_v:" << left_v << " right_v:" << right_v << endl;
    // cout << "left:" << left << " right:" << right << endl;
    // cout << res << endl;
    counter++;
  }

  /*
  if (N%2 == 0) {
    int tmp = res + abs(right_v - A[right]);
    int tmp2 = res + abs(left_v - A[right]);
    if (tmp < tmp2) res = tmp2;
    else res = tmp;
  }
  */

  cout << res << endl;
}