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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long[] a = new long[n];
        for(int i=0;i<n;i++) a[i] = sc.nextLong();
        Arrays.sort(a);
        long sum = 0;
        for(long w:a) sum += w;
        long ans = n%2==0?solveEven(a,sum,n):solveOdd(a,sum,n);
        System.out.println(ans);
    }
    // when n is odd
    static long solveOdd(long[] a, long sum, int n){
        long sum1 = 0;
        for(int i=n/2+1;i<=n-1;i++) sum1 += a[i];
        long ans1 = 2*sum1-2*(sum-sum1)+a[n/2]+a[n/2-1];
        long sum2 = 0;
        for(int i=n/2;i<n;i++) sum2 += a[i];
        long ans2 = 2*sum2-a[n/2]-a[n/2+1]-2*(sum-sum2);
        return Math.max(ans1,ans2);
    }
    static long solveEven(long[] a, long sum, int n){
        if(n==2) return a[1]-a[0];
        long sum1 = 0;
        for(int i=n/2;i<n;i++) sum1 += a[i];
        long ans1 = 2*sum1-a[n/2]-2*(sum-sum1)+a[n/2-1];
        return ans1;
    }
}