Problem:
Score : 700 points

Problem Statement

We have N cards numbered 1, 2, ..., N.
Card i (1 \leq i \leq N) has an integer A_i written in red ink on one side and an integer B_i written in blue ink on the other side.
Initially, these cards are arranged from left to right in the order from Card 1 to Card N, with the red numbers facing up.

Determine whether it is possible to have a non-decreasing sequence facing up from left to right (that is, for each i (1 \leq i \leq N - 1), the integer facing up on the (i+1)-th card from the left is not less than the integer facing up on the i-th card from the left) by repeating the operation below. If the answer is yes, find the minimum number of operations required to achieve it.

Choose an integer i (1 \leq i \leq N - 1).
Swap the i-th and (i+1)-th cards from the left, then flip these two cards.

Constraints

1 \leq N \leq 18

1 \leq A_i, B_i \leq 50 (1 \leq i \leq N)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N
B_1 B_2 ... B_N

Output

If it is impossible to have a non-decreasing sequence, print -1.
If it is possible, print the minimum number of operations required to achieve it.

Sample Input 1

3
3 4 3
3 2 3

Sample Output 1

1

By doing the operation once with i = 1, we have a sequence [2, 3, 3] facing up, which is non-decreasing.

Sample Input 2

2
2 1
1 2

Sample Output 2

-1

After any number of operations, we have the sequence [2, 1] facing up, which is not non-decreasing.

Sample Input 3

4
1 2 3 4
5 6 7 8

Sample Output 3

0

No operation may be required.

Sample Input 4

5
28 15 22 43 31
20 22 43 33 32

Sample Output 4

-1

Sample Input 5

5
4 46 6 38 43
33 15 18 27 37

Sample Output 5

3

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)


// Problem : D - Swap and Flip
// Contest : AtCoder - Keyence Programming Contest 2020
// URL : https://atcoder.jp/contests/keyence2020/tasks/keyence2020_d
// Memory Limit : 1024 MB
// Time Limit : 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

//Macros

#define ll long long
#define db long double
#define pb push_back
#define eb emplace_back
#define pii pair<int,int>
#define vi vector<int>
#define vii vector<pii>
#define mi map<int,int>
#define mii map<pii,int>
#define fi first
#define se second
#define all(a) (a).begin(),(a).end()
#define sz(x) (int)x.size()
#define mod 1000000007
#define EPS 1e-9
#define io ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define inf 1e10
#define PI acos(-1.0)
#define int long long
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define N 20
#define MX 52

int n;
int dp[1LL<<N][MX], a[N], b[N];
int val[N][N];

//Solution

void solve(){
    cin>>n;
    for(int i=0;i<n;i++) cin>>a[i];
    for(int j=0;j<n;j++) cin>>b[j];
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            val[i][j] = (i%2 == j%2 ? a[j] : b[j]);
        }
    }
    for(int i=0;i<=(1<<n);i++){
        for(int j=0;j<MX;j++) dp[i][j] = inf;
    }
    dp[0][0] = 0;
    for(int mask=0;mask<(1<<n);mask++){
        int i = __builtin_popcount(mask);
        for(int num=0;num<MX;num++){
            int cost = 0;
            for(int j=0;j<n;j++){
                if((mask>>j)&1) continue;
                if(val[i][j] >= num)
                    dp[mask|(1<<j)][val[i][j]] = min(dp[mask|(1<<j)][val[i][j]], dp[mask][num]+cost);
                cost++;
            }
        }
    }
    int ans = inf;
    for(int i=0;i<MX;i++) ans = min(ans,dp[(1<<n)-1][i]);
    cout<<(ans == inf ? -1 : ans)<<endl;
}

signed main(){
    io;
    //freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
    int tests;
    //cin>>tests;
    tests = 1;
    while(tests--){
        solve();
    }
    return 0;
}