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: Wrong Answer)

#include <fstream>
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <string>
#include <sstream>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <random>

using namespace std;

#define INT_MAX_VALUE 2147483647
#define LONG_LONG_MAX_VALUE 9223372036854775807
#define ll long long
#define ld long double

struct XX{
    int s;
    int t;
};

class xxGreater {
public:
    bool operator()(const XX& riLeft, const XX& riRight) const {
        //第2条件
        if((riLeft.t) == (riRight.t)){
            return riLeft.s < riRight.s;//<:昇順(小さいものから順番)、>:降順(大きいものから順番)
            //プライオリティキューの場合は > で、top()すると値の小さいものがとれる
        }
        //第1条件
        return (riLeft.t) < (riRight.t);
    }
};


//map<long long,long long> prime_f(long long n){
//    map<long long,long long>res;
//    for(int i=2;i*i<=n;i++){
//        while(n%i==0){
//            ++res[i];
//            n/=i;
//        }
//    }
//    if(n!=1)res[n]=1;
//    return res;
//}

int n;
//int dat[2*10000000];
//int dat2[2*10000000];
int dat[10];
int dat2[10];

void init(int n_){
    n=1;
    while(n<n_)n*=2;
    for(int i=0;i<2*n-1;i++){
        dat[i]=0;
        dat2[i]=0;
    }
}

void initset(int k,int a){
    k+=n-1;
    dat[k]=a;
    while(k>0){
        k=(k-1)/2;
        dat[k]=dat[k*2+1]+dat[k*2+2];
    }
}

//[a,b)の間を[l,r]区間で比較しアップデート
//引数のindexに注意
//nは固定。initで計算すみ
//update2(L[i],R[i]+1,0,0,n,D[i]);
void update2(int a,int b,int k,int l,int r,int v){//v更新値、区間は0-index
    if(r<=a || b<=l)return;
    if(a<=l && r<=b){
        dat[k]+=dat2[k];
        if(r-l>1){
            dat2[k*2+1]+=dat2[k]/2;
            dat2[k*2+1]+=dat2[k]/2;
        }
        dat2[k]=v*(r-l);
        return;
    }else{
        update2(a,b,k*2+1,l,(l+r)/2,v);
        update2(a,b,k*2+2,(l+r)/2,r,v);
        return;
    }
}

int query(int a,int b,int k,int l,int r){
    if(r<=a || b<=l)return 0;
    if(a<=l && r<=b){
        dat[k]+=dat2[k];
        if(r-l>1){
            dat2[k*2+1]+=dat2[k]/2;
            dat2[k*2+1]+=dat2[k]/2;
        }
        dat2[k]=0;
        return dat[k];
    }
    else{
        int vl=query(a,b,k*2+1,l,(l+r)/2);
        int vr=query(a,b,k*2+2,(l+r)/2,r);
        return vl+vr;
    }
}



struct edge{
    int to;
    int cost;
};
typedef pair<int,int> P;//first:最短距離、second:頂点番号
vector<edge>G[100000];//G[各頂点番号]
int d[100000];

void dijkstra(int s,int V){//V:頂点数
    priority_queue<P,vector<P>,greater<P>> que;
    fill(d,d+V+1,INT_MAX_VALUE);
    d[s]=0;
    que.push(P(0,s));
    
    while(!que.empty()){
        P p=que.top();
        que.pop();
        int v=p.second;
        if(d[v]<p.first){
            continue;
        }
        for(int i=0;i<G[v].size();i++){
            edge e=G[v][i];
            if(d[e.to]>d[v]+e.cost){
                d[e.to]=d[v]+e.cost;
                que.push(P(d[e.to],e.to));
            }
        }
    }
}
int main(int argc, const char * argv[])
{
    //scanf("%s",S);
    //scanf("%d",&N);
    //scanf("%lld %lld",&target1,&target2);
    //sscanf(tmp.c_str(),"%dd%d%d",&time[i], &dice[i], &z[i]);
    //getline(cin, target);
    //ifstream ifs("a.txt");//テスト用
    //ifs >> a;
    //ここから
    
    //入力高速化
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    int N;
    cin >> N;
    ll a[100000];
    for(int i=0;i<N;i++){
        cin >> a[i];
    }
    sort(a,a+N);
    ll b[100000];
    int mid=N/2-1;
    b[mid]=a[0];
    for(int i=1;i<N-1;i=i+2){
        if(mid+(2*i)<N)
            b[mid+(2*i)]=a[i];
        if(mid-(2*i)>=0)
            b[mid-(2*i)]=a[i+1];
    }
    sort(a,a+N,greater<ll>());
    mid=N/2;
    b[mid]=a[0];
    for(int i=1;i<N-1;i=i+2){
        if(mid-(2*i)>=0)
            b[mid-(2*i)]=a[i];
        if(mid+(2*i)<N)
            b[mid+(2*i)]=a[i+1];
    }
    ll ans=0;
    for(int j=0;j<N-1;j++){
        ans+=abs(b[j+1]-b[j]);
    }
    ans+=abs(b[N-1]-b[0]);
    ll tmp=2147483647;
    for(int j=0;j<N-1;j++){
        tmp=min(tmp,abs(b[j+1]-b[j]));
    }
    tmp=min(tmp,abs(b[N-1]-b[0]));
    
    cout << ans-tmp << endl;
    
    //ここまで
    //cout << "ans" << endl;
    //printf("%.0f\n",ans);//小数点以下表示なし
    //printf("%.7f\n",p);
    
    return 0;
}