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>
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define fore(i,a) for(auto &i:a)
#define all(x) (x).begin(),(x).end()
//#pragma GCC optimize ("-O3")
using namespace std; void _main(); int main() { cin.tie(0); ios::sync_with_stdio(false); _main(); }
typedef long long ll; const int inf = INT_MAX / 2; const ll infl = 1LL << 60;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a = b; return 1; } return 0; }
//------------------------------------------------------------------------------
//------------------------------------------------------------------------------
void _main()
{
    int N;
    cin >> N;
    vector<int> A(N);
    rep(i, 0, N) cin >> A[i];
    sort(A.begin(), A.end());
    vector<int> B(N * 2, -1);
    B[N] = A[0];
    int front = 1;
    int back = N - 1;
    int left = N;
    int right = N;
    rep(i, 0, N - 1)
    {
        const bool isBack = (i / 2) % 2 == 0;

        if (i % 2 == 0)
        {
            --left;

            if (isBack) B[left] = A[back--];
            else B[left] = A[front++];
        }
        else
        {
            ++right;

            if (isBack) B[right] = A[back--];
            else B[right] = A[front++];
        }
    }

    int sum = 0;
    rep(i, left, right)
    {
        sum += abs(B[i + 1] - B[i]);
    }
    cout << sum << endl;
}