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: Wrong Answer)


import java.util.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int a[] = new int[n];
		long ans = 0;
		for(int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
		}
		
		Arrays.sort(a);
		if(n%2 == 0) {
			for(int i = 0; i < n/2-1; i++) {
				ans -= 2*a[i];
			}
			ans -= a[n/2-1];
			ans += a[n/2];
			for(int i = n/2+1; i < n; i++) {
				ans += 2*a[i];
			}
		}else if(n >= 5){
			int tmp1 = 0;
			int tmp2 = 0;
			for(int i = 0; i < n/2 -1; i++) {
				tmp1 -= 2*a[i];
			}
			tmp1 -= a[n/2 - 1];
			tmp1 -= a[n/2];
			for(int i = n/2+1; i < n; i++) {
				tmp1+= 2*a[i];
			}
			for(int i = 0; i < n/2; i++) {
				tmp2 -= 2*a[i];
			}
			tmp2 -= a[n/2 + 1];
			tmp2 -= a[n/2];
			for(int i = n/2+2; i < n; i++) {
				tmp2+= 2*a[i];
			}
			ans = Math.max(tmp1, tmp2);
		}else {
			int tmp1 = a[2]*2 - a[1] - a[0];
			int tmp2 = a[2] + a[1] - a[0]*2;
			ans = Math.max(tmp1, tmp2);
			
		}
		System.out.println(ans);
		sc.close();
	}
}