<p>You are given a list of&nbsp;<code>preferences</code>&nbsp;for&nbsp;<code>n</code>&nbsp;friends, where <code>n</code> is always <strong>even</strong>.</p>

<p>For each person <code>i</code>,&nbsp;<code>preferences[i]</code>&nbsp;contains&nbsp;a list of friends&nbsp;<strong>sorted</strong> in the <strong>order of preference</strong>. In other words, a friend earlier in the list is more preferred than a friend later in the list.&nbsp;Friends in&nbsp;each list are&nbsp;denoted by integers from <code>0</code> to <code>n-1</code>.</p>

<p>All the friends are divided into pairs.&nbsp;The pairings are&nbsp;given in a list&nbsp;<code>pairs</code>,&nbsp;where <code>pairs[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> denotes <code>x<sub>i</sub></code>&nbsp;is paired with <code>y<sub>i</sub></code> and <code>y<sub>i</sub></code> is paired with <code>x<sub>i</sub></code>.</p>

<p>However, this pairing may cause some of the friends to be unhappy.&nbsp;A friend <code>x</code>&nbsp;is unhappy if <code>x</code>&nbsp;is paired with <code>y</code>&nbsp;and there exists a friend <code>u</code>&nbsp;who&nbsp;is paired with <code>v</code>&nbsp;but:</p>

<ul>
	<li><code>x</code>&nbsp;prefers <code>u</code>&nbsp;over <code>y</code>,&nbsp;and</li>
	<li><code>u</code>&nbsp;prefers <code>x</code>&nbsp;over <code>v</code>.</li>
</ul>

<p>Return <em>the number of unhappy friends</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 4, preferences = [[1, 2, 3], [3, 2, 0], [3, 1, 0], [1, 2, 0]], pairs = [[0, 1], [2, 3]]
<strong>Output:</strong> 2
<strong>Explanation:</strong>
Friend 1 is unhappy because:
- 1 is paired with 0 but prefers 3 over 0, and
- 3 prefers 1 over 2.
Friend 3 is unhappy because:
- 3 is paired with 2 but prefers 1 over 2, and
- 1 prefers 3 over 0.
Friends 0 and 2 are happy.
</pre>

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

<pre>
<strong>Input:</strong> n = 2, preferences = [[1], [0]], pairs = [[1, 0]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> Both friends 0 and 1 are happy.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 4, preferences = [[1, 3, 2], [2, 3, 0], [1, 3, 0], [0, 2, 1]], pairs = [[1, 3], [0, 2]]
<strong>Output:</strong> 4
</pre>

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

<ul>
	<li><code>2 &lt;= n &lt;= 500</code></li>
	<li><code>n</code>&nbsp;is even.</li>
	<li><code>preferences.length&nbsp;== n</code></li>
	<li><code>preferences[i].length&nbsp;== n - 1</code></li>
	<li><code>0 &lt;= preferences[i][j] &lt;= n - 1</code></li>
	<li><code>preferences[i]</code>&nbsp;does not contain <code>i</code>.</li>
	<li>All values in&nbsp;<code>preferences[i]</code>&nbsp;are unique.</li>
	<li><code>pairs.length&nbsp;== n/2</code></li>
	<li><code>pairs[i].length&nbsp;== 2</code></li>
	<li><code>x<sub>i</sub> != y<sub>i</sub></code></li>
	<li><code>0 &lt;= x<sub>i</sub>, y<sub>i</sub>&nbsp;&lt;= n - 1</code></li>
	<li>Each person is contained in <strong>exactly one</strong> pair.</li>
</ul>

class Solution {
public:
    int unhappyFriends(int n, vector<vector<int>>& preferences, vector<vector<int>>& pairs) {
        vector<vector<int>> ranks(n, vector<int>(n,-1));
        for(auto i=0; i<n; i++){
            for(auto j=0; j<n-1; j++){
                ranks[i][preferences[i][j]] = (n-1)-j;
            }
        }
        unordered_set<int> unhappies;
        for(auto i=0; i<n*0.5; i++){
            int x = pairs[i][0], y = pairs[i][1];
            int xScore = ranks[y][x];
            int yScore = ranks[x][y];
            for(auto j=0; j<n*0.5; j++){
                if(j==i)    continue;
                int u = pairs[j][0], v = pairs[j][1];
                int uScore = ranks[v][u];
                int vScore = ranks[u][v];
                if(ranks[x][u]>yScore && ranks[u][x]>vScore){
                    unhappies.insert(x);
                    unhappies.insert(u);
                }
                if(ranks[x][v]>yScore && ranks[v][x]>uScore){
                    unhappies.insert(x);
                    unhappies.insert(v);
                }
                if(ranks[y][u]>xScore && ranks[u][y]>vScore){
                    unhappies.insert(y);
                    unhappies.insert(u);
                }
                if(ranks[y][v]>xScore && ranks[v][y]>uScore){
                    unhappies.insert(y);
                    unhappies.insert(v);
                }
            }
        }
        return unhappies.size();
    }
};
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