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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#define int long long

signed compair(const void *a, const void *b){
	int sub = *(int *)a - *(int *)b;
	if(sub < 0){
		return -1;
	}
	else if(sub == 0){
		return 0;
	}
	else{
		return 1;
	}
}

int ABS(int a){
	return a >= 0 ? a : -a;
}

int min(int a, int b){
	return a <= b ? a : b;
}

signed main(){
	int N, i, m, ans = 0;
	scanf("%lld", &N);
	int *A = (int *)malloc(sizeof(int) * N);
	for(i = 0; i < N; i++){
		scanf("%lld", &A[i]);
	}
	qsort(A, N, sizeof(int), compair);
	if(N % 2 == 0){
		m = A[N / 2];
		for(i = 0; i < N; i++){
			ans += 2 * ABS(A[i] - m);
		}
		ans -= ABS(A[N / 2 - 1] - m) + ABS(A[N / 2] - m);
	}
	else{
		m = A[N / 2];
		for(i = 0; i < N; i++){
			ans += 2 * ABS(A[i] - m);
		}
		ans -= min(ABS(A[N / 2 - 1] - m), ABS(A[N / 2 + 1] - m));
	}
	printf("%lld\n", ans);
	return 0;
}