Minimum Degree of a Connected Trio in a Graph
<p>You are given an undirected graph. You are given an integer <code>n</code> which is the number of nodes in the graph and an array <code>edges</code>, where each <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> indicates that there is an undirected edge between <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code>.</p>

<p>A <strong>connected trio</strong> is a set of <strong>three</strong> nodes where there is an edge between <b>every</b> pair of them.</p>

<p>The <strong>degree of a connected trio</strong> is the number of edges where one endpoint is in the trio, and the other is not.</p>

<p>Return <em>the <strong>minimum</strong> degree of a connected trio in the graph, or</em> <code>-1</code> <em>if the graph has no connected trios.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1700-1799/1761.Minimum%20Degree%20of%20a%20Connected%20Trio%20in%20a%20Graph/images/trios1.png" style="width: 388px; height: 164px;" />
<pre>
<strong>Input:</strong> n = 6, edges = [[1,2],[1,3],[3,2],[4,1],[5,2],[3,6]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> There is exactly one trio, which is [1,2,3]. The edges that form its degree are bolded in the figure above.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1700-1799/1761.Minimum%20Degree%20of%20a%20Connected%20Trio%20in%20a%20Graph/images/trios2.png" style="width: 388px; height: 164px;" />
<pre>
<strong>Input:</strong> n = 7, edges = [[1,3],[4,1],[4,3],[2,5],[5,6],[6,7],[7,5],[2,6]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are exactly three trios:
1) [1,4,3] with degree 0.
2) [2,5,6] with degree 2.
3) [5,6,7] with degree 2.
</pre>

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

<ul>
	<li><code>2 &lt;= n &lt;= 400</code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>1 &lt;= edges.length &lt;= n * (n-1) / 2</code></li>
	<li><code>1 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt;= n</code></li>
	<li><code>u<sub>i </sub>!= v<sub>i</sub></code></li>
	<li>There are no repeated edges.</li>
</ul>
Graph; Enumeration
class Solution {
public:
    int minTrioDegree(int n, vector<vector<int>>& edges) {
        bool g[n][n];
        memset(g, 0, sizeof g);
        int deg[n];
        memset(deg, 0, sizeof deg);
        for (auto& e : edges) {
            int u = e[0] - 1, v = e[1] - 1;
            g[u][v] = g[v][u] = true;
            deg[u]++, deg[v]++;
        }
        int ans = INT_MAX;
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if (g[i][j]) {
                    for (int k = j + 1; k < n; ++k) {
                        if (g[j][k] && g[i][k]) {
                            ans = min(ans, deg[i] + deg[j] + deg[k] - 6);
                        }
                    }
                }
            }
        }
        return ans == INT_MAX ? -1 : ans;
    }
};