1761. Minimum Degree of a Connected Trio in a Graph

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:

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.

Solution

class Solution {
    public int minTrioDegree(int n, int[][] edges) {
        int[] degrees = new int[n + 1];
        int[][] adjMatrix = new int[n + 1][n + 1];
        for (int[] edge : edges) {
            adjMatrix[edge[0]][edge[1]] = 1;
            adjMatrix[edge[1]][edge[0]] = 1;
            degrees[edge[0]]++;
            degrees[edge[1]]++;
        }
        int minTrios = Integer.MAX_VALUE;
        for (int i = 1; i <= n; i++) {
            for (int j = i + 1; j <= n; j++) {
                if (adjMatrix[i][j] == 0) {
                    continue;
                }
                for (int k = j + 1; k <= n; k++) {
                    if (adjMatrix[j][k] == 0 || adjMatrix[i][k] == 0) {
                        continue;
                    }
                    int trioDegree = degrees[i] + degrees[j] + degrees[k] - 6;
                    minTrios = Math.min(minTrios, trioDegree);
                }
            }
        }
        return minTrios == Integer.MAX_VALUE ? -1 : minTrios;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).