description: A bracket sequence is a string containing only characters "(" and ")". A regular bracket sequence (or, shortly, an RBS) is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example:
  
    * bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)"); 
    * bracket sequences ")(", "(" and ")" are not. 
  
  
  
  Let's denote the concatenation of two strings x and y as x+y. For example, "()()" + ")(" = "()())(".
  
  You are given n bracket sequences s_1, s_2, ..., s_n. You can rearrange them in any order (you can rearrange only the strings themselves, but not the characters in them).
  
  Your task is to rearrange the strings in such a way that the string s_1 + s_2 + ... + s_n has as many non-empty prefixes that are RBS as possible.
  
  Input
  
  The first line contains a single integer n (1 ≤ n ≤ 20).
  
  Then n lines follow, the i-th of them contains s_i — a bracket sequence (a string consisting of characters "(" and/or ")". All sequences s_i are non-empty, their total length does not exceed 4 ⋅ 10^5.
  
  Output
  
  Print one integer — the maximum number of non-empty prefixes that are RBS for the string s_1 + s_2 + ... + s_n, if the strings s_1, s_2, ..., s_n can be rearranged arbitrarily.
  
  Examples
  
  Input
  
  
  2
  (
  )
  
  
  Output
  
  
  1
  
  
  Input
  
  
  4
  ()()())
  (
  (
  )
  
  
  Output
  
  
  4
  
  
  Input
  
  
  1
  (())
  
  
  Output
  
  
  1
  
  
  Input
  
  
  1
  )(()
  
  
  Output
  
  
  0
  
  Note
  
  In the first example, you can concatenate the strings as follows: "(" + ")" = "()", the resulting string will have one prefix, that is an RBS: "()".
  
  In the second example, you can concatenate the strings as follows: "(" + ")" + "()()())" + "(" = "()()()())(", the resulting string will have four prefixes that are RBS: "()", "()()", "()()()", "()()()()".
  
  The third and the fourth examples contain only one string each, so the order is fixed.
solution: #include <bits/stdc++.h>
  using namespace std;
  const int mx = 4e5 + 10;
  int main() {
    int n;
    cin >> n;
    vector<string> a(n);
    for (string& x : a) cin >> x;
    vector<int> bal(n), mn(n);
    vector<vector<int>> occ(n), nocc(n);
    for (int i = 0; i < n; i++) {
      int b = 0, mn0 = 1e9;
      occ[i].resize(mx * 2);
      nocc[i].resize(mx * 2);
      int mini = 1e9;
      for (char x : a[i]) {
        if (x == ')')
          b--;
        else
          b++;
        mini = min(b, mini);
        mn0 = min(mn0, b);
        occ[i][b + mx]++;
        if (b == mini) nocc[i][b + mx]++;
      }
      bal[i] = b, mn[i] = mn0;
    }
    vector<vector<int>> answer((1 << n), vector<int>(2, -1e9));
    answer[0] = {0, 0};
    vector<int> sum(1 << n);
    for (int mask = 0; mask < (1 << n); mask++) {
      for (int i = 0; i < n; i++) {
        if ((1 << i) & mask) {
          sum[mask] = sum[mask ^ (1 << i)] + bal[i];
          break;
        }
      }
      for (int i = 0; i < n; i++) {
        if ((1 << i) & mask) {
          int nmask = (1 << i) ^ mask;
          if (sum[nmask] + mn[i] < 0) {
            answer[mask][1] =
                max(answer[nmask][0] + nocc[i][mx - sum[nmask]], answer[mask][1]);
          } else {
            answer[mask][0] =
                max(answer[mask][0], answer[nmask][0] + occ[i][mx - sum[nmask]]);
          }
        }
      }
    }
    int ans = 0;
    for (auto x : answer) ans = max({x[0], x[1], ans});
    cout << ans;
    return 0;
  }
  