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;

#ifdef LOCAL
#define debug(var) cerr << (#var) << " = " << (var) << endl;
#else
#define debug(var)
#endif

void init() {
  ios_base::sync_with_stdio(0);
  cin.tie(0); cout.tie(0);
}

const int N = 1e5+23;

int a[N];

void solve() {
  int n; scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", a+i);
  }
  sort(a, a+n);
  int left = a[0], right = a[0];
  long long sum = 0;
  for (int i = 1, j = n-1, left = a[0], right = a[0];;) {
    sum += abs(a[j]-left);
    left = a[j--];
    if (i > j) break;
    sum += abs(a[j]-right);
    right = a[j--];
    if (i > j) break;
    sum += abs(a[i]-left);
    left = a[i++];
    if (i > j) break;
    sum += abs(a[i]-right);
    right = a[i++];
    if (i > j) break;
  }
  long long ans = sum; sum = 0;
  for (int i = 1, j = n-1, left = a[n-1], right = a[n-1];;) {
    sum += abs(a[i]-left);
    left = a[i++];
    if (i > j) break;
    sum += abs(a[i]-right);
    right = a[i++];
    if (i > j) break;
    sum += abs(a[j]-left);
    left = a[j--];
    if (i > j) break;
    sum += abs(a[j]-right);
    right = a[j--];
    if (i > j) break;
  }
  printf("%lld", max(sum, ans));
}

int main() {
  init();
  int t = 1; //scanf("%d", &t);
  while (t--) {
    solve();
  }
  return 0;
}