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.*;
public 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);
		if(n==3){
			System.out.println(Math.max(a[2]*2-a[0]-a[1], a[1]+a[2]-a[0]*2));
		}
		else if(n%2==1){		
			long sum1 = 0L;
			long sum2 = 0L;
			for(int i=0; i<=(n-3)/2; i++){
				sum1 = sum1 - 2*a[i] + 2*a[i+(n+1)/2];
				sum2 = sum2 - 2*a[i] + 2*a[i+(n-1)/2];
			}
			sum1 = sum1 - a[(n-1)/2] + a[(n-3)/2];
			sum2 = sum2 + 2*a[n-1] - a[(n-1)/2] - a[(n+1)/2];
			System.out.println(Math.max(sum1, sum2));
		}
		else{
			long sum = 0;
			for(int i=0; i<=n/2-1; i++){
				sum = sum - 2*a[i] + 2*a[i+n/2];
			}
			sum = sum + a[n/2-1] - a[n/2];
			System.out.println(sum);
		}
	}
}