Minimum Degree of a Connected Trio in a Graph

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:

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:

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.

思路： 用connected matrix 去build graph edge check. 然后就是暴力搜索a, b, c 。 找到之后，计算indegree，取最小的就可以了。

class Solution {
    public int minTrioDegree(int n, int[][] edges) {
        int[] indegree = new int[n + 1];
        int[][] matrix = new int[n + 1][n + 1];
        List<List<Integer>> adjlist = new ArrayList<List<Integer>>();
        for(int i = 0; i < n + 1; i++) {
            adjlist.add(new ArrayList<Integer>());
        }
        for(int[] edge: edges) {
            int u = edge[0];
            int v = edge[1];
            
            indegree[u]++;
            indegree[v]++;
            
            matrix[u][v] = 1;
            matrix[v][u] = 1;
            
            adjlist.get(u).add(v);
            adjlist.get(v).add(u);
        }
        
        int res = Integer.MAX_VALUE;
        for(int a = 1; a <= n; a++) {
            for(int i = 0; i < adjlist.get(a).size(); i++) {
                for(int j = i + 1; j < adjlist.get(a).size(); j++) {
                    int b = adjlist.get(a).get(i);
                    int c = adjlist.get(a).get(j);
                    if(matrix[b][c] == 1) {
                        res = Math.min(res, indegree[a] + indegree[b] + indegree[c] - 6);
                    }
                }
            }
        }
        return res == Integer.MAX_VALUE ? -1: res;
    }
}