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;

int main(int argc, char *argv[]) {
  int N;
  cin >> N;
  vector<int> A;
  for(int i = 0; i < N; i++) {
    int t;
    cin >> t;
    A.push_back(t);
  }
  sort(A.begin(), A.end());

  vector<int> B;
  int count = 0;
  while(true) {
    B.push_back(A.back());
    A.pop_back();
    count++;
    if(count == N) break;
    B.push_back(A.front());
    A.erase(A.begin());
    count++;
    if(count == N) break;
  }

  int sum = 0;
  int tmp_sum = 0;
  for(int i = 0; i < N-1; i++) {
    tmp_sum += abs(B[i+1] - B[i]);
  }
  if(tmp_sum > sum) {
    sum = tmp_sum;
  }
  for(int i = 0; i < N; i++) {
    tmp_sum = tmp_sum - abs(B[N-1] - B[N-2]) + abs(B[0] - B[N-1]);
    if(tmp_sum > sum) {
      sum = tmp_sum;
    }
    B.insert(B.begin(), B.back());
    B.pop_back();
  }
  cout << sum << endl;
  return 0;
}