feat: solve No.1761

Author: BaffinLee

1701-1800/1761. Minimum Degree of a Connected Trio in a Graph.md

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+# 1761. Minimum Degree of a Connected Trio in a Graph
+
+- Difficulty: Hard.
+- Related Topics: Graph.
+- Similar Questions: Add Edges to Make Degrees of All Nodes Even.
+
+## Problem
+
+You are given an undirected graph. You are given an integer `n` which is the number of nodes in the graph and an array `edges`, where each `edges[i] = [ui, vi]` indicates that there is an undirected edge between `ui` and `vi`.
+
+A **connected trio** is a set of **three** nodes where there is an edge between **every** pair of them.
+
+The **degree of a connected trio** is the number of edges where one endpoint is in the trio, and the other is not.
+
+Return **the **minimum** degree of a connected trio in the graph, or** `-1` **if the graph has no connected trios.**
+
+ 
+Example 1:
+
+![](https://assets.leetcode.com/uploads/2021/01/26/trios1.png)
+
+```
+Input: n = 6, edges = [[1,2],[1,3],[3,2],[4,1],[5,2],[3,6]]
+Output: 3
+Explanation: There is exactly one trio, which is [1,2,3]. The edges that form its degree are bolded in the figure above.
+```
+
+Example 2:
+
+![](https://assets.leetcode.com/uploads/2021/01/26/trios2.png)
+
+```
+Input: n = 7, edges = [[1,3],[4,1],[4,3],[2,5],[5,6],[6,7],[7,5],[2,6]]
+Output: 0
+Explanation: 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.
+```
+
+ 
+**Constraints:**
+
+
+	
+- `2 <= n <= 400`
+	
+- `edges[i].length == 2`
+	
+- `1 <= edges.length <= n * (n-1) / 2`
+	
+- `1 <= ui, vi <= n`
+	
+- `ui != vi`
+	
+- There are no repeated edges.
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number} n
+ * @param {number[][]} edges
+ * @return {number}
+ */
+var minTrioDegree = function(n, edges) {
+    var { map, lenMap } = buildMap(n, edges);
+    var res = Number.MAX_SAFE_INTEGER;
+    for (var i = 0; i < map.length; i++) {
+        var arr = Object.keys(map[i]);
+        for (var j = 0; j < arr.length; j++) {
+            var m = arr[j];
+            if (m < i) continue;
+            for (var k = j + 1; k < arr.length; k++) {
+                var n = arr[k];
+                if (n < i) continue;
+                if (map[m][n]) {
+                    res = Math.min(
+                        res,
+                        arr.length - 2 + lenMap[m] - 2 + lenMap[n] - 2,
+                    );
+                }
+            }
+        }
+    }
+    return res === Number.MAX_SAFE_INTEGER ? -1 : res;
+};
+
+var buildMap = function(n, edges) {
+    var map = Array(n + 1).fill(0).map(() => ({}));
+    var lenMap = Array(n + 1).fill(0);
+    for (var i = 0; i < edges.length; i++) {
+        var [m, n] = edges[i];
+        map[m][n] = 1;
+        map[n][m] = 1;
+        lenMap[n]++;
+        lenMap[m]++;
+    }
+    return { map, lenMap };
+};
+```
+
+**Explain:**
+
+Brute-force with hashmap.
+
+**Complexity:**
+
+* Time complexity : O(n ^ 3).
+* Space complexity : O(n ^ 2).