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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn=1e5+100;
int a[maxn];
int main(){
	int n;
	while(~scanf("%d",&n)){
		for(int i=1;i<=n;i++)
			scanf("%d",a+i);
		sort(a+1,a+n+1);
		long long ans=0;
		if(n<=3){
			if(n==2)
				ans=a[2]-a[1];
			else
				ans=max(a[2]+a[3]-2*a[1],2*a[3]-a[1]-a[2]);
			goto loop;
		}
		if(n&1){
			for(int i=1;i<=n;i++){
				if(i==n/2+1)continue;
				else if(i<n/2+1){
					if(i==n/2)
						ans-=a[i];
					else
						ans-=a[i]*2;
				}
				else if(i>n/2+1){
					if(i==n/2+2)
						ans+=a[i];
					else
						ans+=a[i]*2;
				}
			}
			ans+=max(a[n/2+1]-a[n/2],a[n/2+2]-a[n/2+1]);
		}
		else{
			for(int i=1;i<=n;i++){
				if(i<n/2)
					ans-=a[i]*2;
				else if(i>n/2+1)
					ans+=a[i]*2;			
			}
			ans+=a[n/2+1]-a[n/2];
		}
loop:	printf("%lld\n",ans);
	}
}