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 Rust (1.42.0) (judge verdict: Accepted)

use std::io::Read;
use std::cmp::max;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();

    let answer = solve(&buf);

    println!("{}", answer);
}

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    let n: usize = iterator.next().unwrap().parse().unwrap();
    let mut a: Vec<isize> = (0..n).map(|_| iterator.next().unwrap().parse().unwrap()).collect();
    a.sort();

    if n % 2 == 0 {
        let mut b: Vec<isize> = vec![-1, 1];
        for _ in 0..((n - 2) / 2) {
            b.push(2);
            b.push(-2);
        }
        b.sort();

        let mut ans = 0;
        for i in 0..n {
            ans += a[i] * b[i];
        }
        return ans.to_string();
    }

    let mut b1: Vec<isize> = vec![-1, -1];
    let mut b2: Vec<isize> = vec![1, 1];
    for _ in 0..(n / 2) {
        b1.push(2);
        b2.push(-2);
    }
    for _ in 0..(n / 2 - 1) {
        b1.push(-2);
        b2.push(2);
    }
    b1.sort();
    b2.sort();

    let mut ans1 = 0;
    let mut ans2 = 0;
    for i in 0..n {
        ans1 += a[i] * b1[i];
        ans2 += a[i] * b2[i];
    }
    max(ans1, ans2).to_string()
}