<p>On a social network consisting of <code>m</code> users and some friendships between users, two users can communicate with each other if they know a common language.</p>

<p>You are given an integer <code>n</code>, an array <code>languages</code>, and an array <code>friendships</code> where:</p>

<ul>
	<li>There are <code>n</code> languages numbered <code>1</code> through <code>n</code>,</li>
	<li><code>languages[i]</code> is the set of languages the <code>i<sup>​​​​​​th</sup></code>​​​​ user knows, and</li>
	<li><code>friendships[i] = [u<sub>​​​​​​i</sub>​​​, v<sub>​​​​​​i</sub>]</code> denotes a friendship between the users <code>u<sup>​​​​​</sup><sub>​​​​​​i</sub></code>​​​​​ and <code>v<sub>i</sub></code>.</li>
</ul>

<p>You can choose <strong>one</strong> language and teach it to some users so that all friends can communicate with each other. Return <i data-stringify-type="italic">the</i> <i><strong>minimum</strong> </i><i data-stringify-type="italic">number of users you need to teach.</i></p>
Note that friendships are not transitive, meaning if <code>x</code> is a friend of <code>y</code> and <code>y</code> is a friend of <code>z</code>, this doesn&#39;t guarantee that <code>x</code> is a friend of <code>z</code>.
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 2, languages = [[1],[2],[1,2]], friendships = [[1,2],[1,3],[2,3]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> You can either teach user 1 the second language or user 2 the first language.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 3, languages = [[2],[1,3],[1,2],[3]], friendships = [[1,4],[1,2],[3,4],[2,3]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> Teach the third language to users 1 and 3, yielding two users to teach.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n &lt;= 500</code></li>
	<li><code>languages.length == m</code></li>
	<li><code>1 &lt;= m &lt;= 500</code></li>
	<li><code>1 &lt;= languages[i].length &lt;= n</code></li>
	<li><code>1 &lt;= languages[i][j] &lt;= n</code></li>
	<li><code>1 &lt;= u<sub>​​​​​​i</sub> &lt; v<sub>​​​​​​i</sub> &lt;= languages.length</code></li>
	<li><code>1 &lt;= friendships.length &lt;= 500</code></li>
	<li>All tuples <code>(u<sub>​​​​​i, </sub>v<sub>​​​​​​i</sub>)</code> are unique</li>
	<li><code>languages[i]</code> contains only unique values</li>
</ul>

#include <iostream>
#include <random>
#include <array>
#include <vector>
#include <string>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <algorithm>
#include <functional>
#include <queue>
#include <stack>
#include <bitset>

#include <math.h>

using namespace std;

using LL = long long;

int const INF = numeric_limits<int>::max();

class Solution {
  public:
    int minimumTeachings(int n, vector<vector<int>>& langs, vector<vector<int>>& friends) {
      int m = langs.size();
      int ans = INF;

      vector<pair<int, int>> others;

      for (auto& connection: friends) {
        int first = connection[0] - 1;
        int second = connection[1] - 1;

        unordered_set<int> firstKnows(langs[first].begin(), langs[first].end());

        bool ok = false;

        for (int lang: langs[second]) {
          if (firstKnows.contains(lang)) {
            ok = true;
            break;
          }
        }

        if (!ok) {
          others.emplace_back(first, second);
        }
      }

      vector<unordered_set<int>> knows(m);

      for (int user = 0; user < m; ++user) {
        for (int lang: langs[user]) {
          knows[user].insert(lang - 1);
        }
      }

      unordered_set<int> cnt;

      for (int lang = 0; lang < n; ++lang) {
        for (auto&&[first, second]: others) {
          if (!knows[first].contains(lang)) cnt.insert(first);
          if (!knows[second].contains(lang)) cnt.insert(second);
        }

        ans = min(ans, static_cast<int>(cnt.size()));

        cnt.clear();
      }

      return ans;
    }
};

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