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 max(a,b) ((a)>(b)?(a):(b))

int compare_int(const void *a,const void *b){
	return *(long long*)a-*(long long*)b;
}

int main(){
	int n,i;
	long long a[100000],ans=0;
	scanf("%d",&n);
	for(i=0;i<n;i++)scanf("%lld",&a[i]);
	qsort(a,n,sizeof(long long),compare_int);
	if(n%2==0){
		for(i=0;i<n;i++){
			if(i<n/2){
				if(i==n/2-1)ans-=a[i];
				else ans-=2*a[i];
			}
			else{
				if(i==n/2)ans+=a[i];
				else ans+=2*a[i];
			}
		}
	}else{
		long long ans2=0;
		for(i=0;i<n;i++){
			if(i<n/2){
				ans-=2*a[i];
			}
			else{
				if(i<=n/2+1)ans+=a[i];
				else ans+=2*a[i];
			}
		}
		for(i=0;i<n;i++){
			if(i<=n/2){
				if(i>=n/2-1)ans2-=a[i];
				else ans2-=2*a[i];
			}
			else{
				ans2+=2*a[i];
			}
		}
		ans=max(ans,ans2);
	}
	printf("%lld\n",ans);
	return 0;
}