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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;

public class Main {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		int n = Integer.parseInt(br.readLine());

		ArrayList<Long> pq1 = new ArrayList<>();
		ArrayList<Long> pq2 = new ArrayList<>();

		for (int i = 0; i < n; i++) {
			Long a = Long.parseLong(br.readLine());

			pq1.add(a);
			pq2.add(a);
		}

		Collections.sort(pq1);
		Collections.sort(pq2, Comparator.reverseOrder());

		long left = pq1.get(0);
		long right = pq2.get(0);

		int one = 1;
		int two = 1;

		long ans = Math.abs(left - right);

		for (int i = 2; i < n; i++) {
			long num = 0;
			if (i % 2 == 0) {
				num = pq1.get(one++);
			} else {
				num = pq2.get(two++);
			}
			if (Math.abs(left - num) > Math.abs(right - num)) {
				ans += Math.abs(left - num);
				left = num;
			} else {
				ans += Math.abs(right - num);
				right = num;
			}
		}

		long ans2 = -1;

		if (n % 2 != 0) {

			left = pq1.get(0);
			right = pq2.get(0);

			one = 1;
			two = 1;

			ans2 = Math.abs(right - left);

			for (int i = 2; i < n; i++) {
				long num = 0;
				if (i % 2 == 0) {
					num = pq2.get(two++);
				} else {
					num = pq1.get(one++);
				}
				if (Math.abs(left - num) > Math.abs(right - num)) {
					ans2 += Math.abs(left - num);
					left = num;
				} else {
					ans2 += Math.abs(right - num);
					right = num;
				}
			}
		}

		System.out.println(Math.max(ans, ans2));
	}
}