https://codeforces.com/contest/1573/problem/B
#include<iostream>
#include<algorithm>
#include<bits/stdc++.h>
using namespace std;

const int N = 1e3 + 10;
typedef long long ll;
typedef pair<char, int > PII;
 
 

int n;

int cmp (vector<int> a, vector<int> b) {
  
}
void solve() {
  cin >> n;
  vector<int> a(n * 2 + 1), b(n * 2 + 1);
  int minn = 0x3f3f3f3f;
  for (int i = 1;  i<= n; i++) {
    int t;
    cin >>t;
    a[t] = i;
  }
  for (int i =1; i<= n; i ++) {
    int t;
    cin >> t;
    b[t] = i;
  }
  int ans =0x3f3f3f3f;
  for (int i = 2; i <=n * 2; i += 2) {
    minn = min (minn, a[i -1]);
    ans = min (ans, minn + b[i] - 2);
  } 
  cout << ans << endl;
  
}
int main () {
  int t;
  cin >> t;
  while (t --) solve();
  return 0;
}

只要第一个位置满足条件就行了。为此我们从小到大遍历，然后选择坐标最小的，并且始终满足条件。