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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <cmath>
#include <cstring>
#include <utility>
#include <numeric>

using namespace std;

const int INF = 1e9;

int main(){

    int N;
    cin >> N;
    vector<int> A(N), B(N);
    for(int i = 0; i < N; i++){
        cin >> A[i];
    }
    for(int j = 0; j < N; j++){
        cin >> B[j];
    }

    int ans = INF;
    int com = pow(2, N);
    for(int i = 0; i < com; i++){
        int sum = 0;
        vector<pair<int, int>> v(N);
        for(int j = 0; j < N; j++){
            if(i & (1 << j)){
                sum++;
                v[j] = make_pair(B[j], j);
            }
            else{
                v[j] = make_pair(A[j], j);
            }
        }

        if(sum % 2 == 1){
            continue;
        }

        sort(v.begin(), v.end());
        int num = 0;
        vector<int> cnt(N, 0);

        for(int j = 0; j < N; j++){
            for(int k = j + 1; k < N; k++){
                if(v[j].second > v[k].second){
                    num++;
                    cnt[v[j].second]++;
                    cnt[v[k].second]++;
                }
            }
        }

        for(int j = 0; j < N; j++){
            cnt[j] = (cnt[j] % 2);
        }

        vector<int> cnt1(51, 0);
        for(int j = 0; j < N; j++){
            if(i & (1 << j)){
                cnt1[B[j]] += cnt[j];
            }
            else{
                cnt1[A[j]] += cnt[j];
            }
        }

        vector<int> cnt2(51, 0);
        for(int j = 0; j < N; j++){
            if(i & (1 << j)){
                cnt2[B[j]] += 1;
            }
        }

        bool isOk = true;
        for(int j = 0; j < 51; j++){
            if(cnt1[j] != cnt2[j]){
                isOk = false;
                break;
            }
        }

        if(isOk){
            vector<vector<pair<int, int>>> v01(51);
            for(int j = 0; j < N; j++){
                if(i & (1 << j)){
                    v01[B[j]].push_back(make_pair(cnt[j], 1));
                }
                else{
                    v01[A[j]].push_back(make_pair(cnt[j], 0));
                }
            }

            int cost = 0;
            for(int j = 0; j < 51; j++){
                int vSize = (int)v01[j].size();
                int indexFirst = 0;
                int indexSecond = 0;
                while(indexFirst < vSize && indexSecond < vSize){
                    if(v01[j][indexFirst].first == 1 && v01[j][indexSecond].second == 1){
                        cost += abs(indexSecond - indexFirst);
                        indexFirst++;
                        indexSecond++;
                    }
                    else if(v01[j][indexFirst].first == 1){
                        indexSecond++;
                    }
                    else{
                        indexFirst++;
                    }
                }
            }

            ans = min(ans, num + cost);
        }
    }

    if(ans == INF){
        cout << -1 << endl;
    }
    else{
        cout << ans << endl;
    }

    return 0;
}