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];
		int[] B = new int[N];
		
		for(int i = 0; i < N; i++) {
			A[i] = sc.nextInt();
		}
		
		Arrays.sort(A);
		
		int a = 0;
		int b = 0;
		
		if(N % 2 == 0) {
			B[N / 2] = A[0];
			for(int i = 1; i < N; i++) {				
				if((i % 2 != 0) && (a == 0)) {
					B[(N / 2) - (i + 1) / 2] = A[N - (i + 1) / 2];
					a = 1;
				} else if((i % 2 != 0) && (a == 1)) {
					B[(N / 2) - (i + 1) / 2] = A[(i + 1) / 2 - 1];
					a = 0;		
				} else if((i % 2 == 0) && (b == 0)){
					B[(N / 2) + (i / 2)] = A[N - (i + 2) / 2];
					b = 1;
				} else if((i % 2 == 0) && (b == 1)){
					B[(N / 2) + (i / 2)] = A[(i + 2) / 2 - 1];
					b = 0;					
				}
			}
		}
		
		if(N % 2 != 0) {
			B[(N - 1) / 2] = A[0];
			for(int i = 1; i < N; i++) {				
				if((i % 2 != 0) && (a == 0)) {
					B[(N - 1) / 2 - (i + 1) / 2] = A[N - (i + 1) / 2];
					a = 1;
				} else if((i % 2 != 0) && (a == 1)) {
					B[(N - 1) / 2 - (i + 1) / 2] = A[(i + 1) / 2 - 1];
					a = 0;		
				} else if((i % 2 == 0) && (b == 0)){
					B[(N - 1) / 2 + (i / 2)] = A[N - (i + 2) / 2];
					b = 1;
				} else if((i % 2 == 0) && (b == 1)){
					B[(N - 1) / 2 + (i / 2)] = A[(i + 2) / 2 - 1];
					b = 0;					
				}
			}	
		}
		
		int diff = 0;
		int sum = 0;
		
		for(int i = 1; i < N; i++) {
			diff = Math.abs(B[i] - B[i - 1]);
			sum = sum + diff;
		}
		

		System.out.println(sum);

	}
}