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) {
Map<Integer, Integer> degree = new HashMap();
boolean[][] matrix = new boolean[n + 1][n + 1];
for (int[] edge : edges) {
degree.put(edge[0], degree.getOrDefault(edge[0], 0) + 1);
degree.put(edge[1], degree.getOrDefault(edge[1], 0) + 1);
matrix[edge[0]][edge[1]] = true;
matrix[edge[1]][edge[0]] = true;
}
int min = Integer.MAX_VALUE;
for (int[] edge : edges) {
int i = edge[0], j = edge[1];
for (int k = 1; k <= n; k ++) {
if (matrix[i][k] && matrix[j][k]) {
min = Math.min(min, degree.get(i) + degree.get(j) + degree.get(k) - 6);
}
}
}
return min == Integer.MAX_VALUE ? -1 : min;
}
}