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-1if the graph has no connected trios.
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.
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.
2 <= n <= 400edges[i].length == 21 <= edges.length <= n * (n-1) / 21 <= ui, vi <= nui != vi- There are no repeated edges.
use std::collections::HashSet;implSolution{pubfnmin_trio_degree(n:i32,edges:Vec<Vec<i32>>) -> i32{let n = n asusize;letmut edges_set = HashSet::new();letmut count = vec![0; n + 1];letmut ret = -1;for i in0..edges.len(){let(u, v) = ( edges[i][0].min(edges[i][1])asusize, edges[i][0].max(edges[i][1])asusize,); edges_set.insert((u, v)); count[u] += 1; count[v] += 1;}for i in1..=n {for j in i + 1..=n {if !edges_set.contains(&(i, j)){continue;}for k in j + 1..=n {if !edges_set.contains(&(i, k)) || !edges_set.contains(&(j, k)){continue;}let degree = count[i] + count[j] + count[k] - 6;if ret == -1 || degree < ret { ret = degree;}}}} ret }}