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: Accepted)

#include <bits/stdc++.h>

using namespace std;

int a[20], b[20];
int dp[1 << 19][20];

int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	
	int n;
	cin >> n;
	for (int i = 0; i < n; i++) cin >> a[i];
	for (int i = 0; i < n; i++) cin >> b[i];
	memset(dp, 255, sizeof(dp));
	
	for (int i = 0; i < n; i++) dp[1 << i][i] = 0;
	for (int mask = 1; mask < (1 << n); mask++) {
		int c = 0;
		for (int i = 0; i < n; i++) {
			if (mask >> i & 1) c++;
		}
		for (int i = 0; i < n; i++) {
			if (!(mask >> i & 1) || dp[mask][i] == -1) continue;
			int w = (i % 2 == c % 2 ? b[i] : a[i]);
			int cost = c;
			for (int j = 0; j < n; j++) {
				if (mask >> j & 1) cost--;
				else { 
					int v = (j % 2 == c % 2 ? a[j] : b[j]);
					if (w <= v) {
						int val = dp[mask][i] + cost;
						if (dp[mask | 1 << j][j] == -1 || dp[mask | 1 << j][j] > val) {
							dp[mask | 1 << j][j] = val;
						}
					}
				}
			}
		}
	}
	
	int res = -1;
	for (int i = 0; i < n; i++) {
		if (dp[(1 << n) - 1][i] != -1) {
			if (res == -1 || res > dp[(1 << n) - 1][i]) {
				res = dp[(1 << n) - 1][i];
			}
		}
	}
	
	cout << res << '\n';
}