feat: add solution 1761. Minimum Degree of a Connected Trio in a Graph

Author: tzuyi0817

solutions/1061. Lexicographically Smallest Equivalent String/README.md

@@ -64,7 +64,7 @@ So only the second letter 'o' in baseStr is changed to 'd', the answer is "hdld"
 **Solution: `Union Find`**
 
 - Time complexity: <em>O(m+n)</em>
-- Space complexity: <em>O(26m -> m)</em>
+- Space complexity: <em>O(26+m -> m)</em>
 
 <p>&nbsp;</p>
 

solutions/1761. Minimum Degree of a Connected Trio in a Graph/README.md

@@ -0,0 +1,90 @@
+# [1761. Minimum Degree of a Connected Trio in a Graph](https://leetcode.com/problems/minimum-degree-of-a-connected-trio-in-a-graph)
+
+## Description
+
+<div class="elfjS" data-track-load="description_content"><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://assets.leetcode.com/uploads/2021/01/26/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://assets.leetcode.com/uploads/2021/01/26/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>
+</div>
+
+<p>&nbsp;</p>
+
+## Solutions
+
+**Solution: `Brute Force`**
+
+- Time complexity: <em>O(n<sup>3</sup>)</em>
+- Space complexity: <em>O(n<sup>2</sup>)</em>
+
+<p>&nbsp;</p>
+
+### **JavaScript**
+
+```js
+/**
+ * @param {number} n
+ * @param {number[][]} edges
+ * @return {number}
+ */
+const minTrioDegree = function (n, edges) {
+  const indegree = Array.from({ length: n + 1 }, () => 0);
+  const connected = Array.from({ length: n + 1 }, () => new Array(n + 1).fill(false));
+  let result = Number.MAX_SAFE_INTEGER;
+
+  for (const [u, v] of edges) {
+    indegree[u] += 1;
+    indegree[v] += 1;
+    connected[u][v] = true;
+    connected[v][u] = true;
+  }
+
+  for (let u = 1; u <= n - 2; u++) {
+    for (let v = u + 1; v <= n - 1; v++) {
+      if (!connected[u][v]) continue;
+
+      for (let k = v + 1; k <= n; k++) {
+        if (!connected[v][k] || !connected[u][k]) continue;
+        const degree = indegree[u] + indegree[v] + indegree[k] - 6;
+
+        result = Math.min(degree, result);
+      }
+    }
+  }
+
+  return result === Number.MAX_SAFE_INTEGER ? -1 : result;
+};
+```

solutions/1761. Minimum Degree of a Connected Trio in a Graph/solution.js

@@ -0,0 +1,32 @@
+/**
+ * @param {number} n
+ * @param {number[][]} edges
+ * @return {number}
+ */
+const minTrioDegree = function (n, edges) {
+  const indegree = Array.from({ length: n + 1 }, () => 0);
+  const connected = Array.from({ length: n + 1 }, () => new Array(n + 1).fill(false));
+  let result = Number.MAX_SAFE_INTEGER;
+
+  for (const [u, v] of edges) {
+    indegree[u] += 1;
+    indegree[v] += 1;
+    connected[u][v] = true;
+    connected[v][u] = true;
+  }
+
+  for (let u = 1; u <= n - 2; u++) {
+    for (let v = u + 1; v <= n - 1; v++) {
+      if (!connected[u][v]) continue;
+
+      for (let k = v + 1; k <= n; k++) {
+        if (!connected[v][k] || !connected[u][k]) continue;
+        const degree = indegree[u] + indegree[v] + indegree[k] - 6;
+
+        result = Math.min(degree, result);
+      }
+    }
+  }
+
+  return result === Number.MAX_SAFE_INTEGER ? -1 : result;
+};