Formatted question description: https://leetcode.ca/all/1761.html

# 1761. Minimum Degree of a Connected Trio in a Graph

## Level

Hard

## Description

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] = [u_i, v_i]`

indicates that there is an undirected edge between `u_i`

and `v_i`

.

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 <= u_i, v_i <= n`

`u_i != v_i`

- There are no repeated edges.

## Solution

Use a hash map to store each node’s connected nodes. Then loop over all tuples of nodes and determine whether the three nodes form a connected trio. If the three nodes form a connected trio, calculate the sum of the numbers of edges that connect to the three nodes and subtract 6 from the sum (to exclude the edges inside the trio). During the process, maintain the minimum degree. Finally, return the minimum degree.

```
class Solution {
public int minTrioDegree(int n, int[][] edges) {
Map<Integer, Set<Integer>> map = new HashMap<Integer, Set<Integer>>();
for (int[] edge : edges) {
int node0 = edge[0], node1 = edge[1];
Set<Integer> set0 = map.getOrDefault(node0, new HashSet<Integer>());
Set<Integer> set1 = map.getOrDefault(node1, new HashSet<Integer>());
set0.add(node1);
set1.add(node0);
map.put(node0, set0);
map.put(node1, set1);
}
int minDegree = Integer.MAX_VALUE;
for (int i = 1; i <= n; i++) {
Set<Integer> set = map.getOrDefault(i, new HashSet<Integer>());
if (set.size() < 2)
continue;
List<Integer> list = new ArrayList<Integer>(set);
int size = list.size();
for (int j = 0; j < size; j++) {
int node1 = list.get(j);
Set<Integer> set1 = map.get(node1);
for (int k = j + 1; k < size; k++) {
int node2 = list.get(k);
if (set1.contains(node2)) {
Set<Integer> set2 = map.get(node2);
int count = set.size() + set1.size() + set2.size() - 6;
minDegree = Math.min(minDegree, count);
}
}
}
}
return minDegree == Integer.MAX_VALUE ? -1 : minDegree;
}
}
```