<p>There is a <strong>bi-directional </strong>graph with <code>n</code> vertices, where each vertex is labeled from <code>0</code> to <code>n - 1</code>. The edges in the graph are represented by a given 2D integer array <code>edges</code>, where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> denotes an edge between vertex <code>u<sub>i</sub></code> and vertex <code>v<sub>i</sub></code>. Every vertex pair is connected by at most one edge, and no vertex has an edge to itself.</p>

<p>Return <em>the length of the <strong>shortest </strong>cycle in the graph</em>. If no cycle exists, return <code>-1</code>.</p>

<p>A cycle is a path that starts and ends at the same node, and each edge in the path is used only once.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/01/04/cropped.png" style="width: 387px; height: 331px;" />
<pre>
<strong>Input:</strong> n = 7, edges = [[0,1],[1,2],[2,0],[3,4],[4,5],[5,6],[6,3]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The cycle with the smallest length is : 0 -&gt; 1 -&gt; 2 -&gt; 0 
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/01/04/croppedagin.png" style="width: 307px; height: 307px;" />
<pre>
<strong>Input:</strong> n = 4, edges = [[0,1],[0,2]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> There are no cycles in this graph.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n &lt;= 1000</code></li>
	<li><code>1 &lt;= edges.length &lt;= 1000</code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>0 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt; n</code></li>
	<li><code>u<sub>i</sub> != v<sub>i</sub></code></li>
	<li>There are no repeated edges.</li>
</ul>

class Solution:
    def findShortestCycle(self, n: int, edges: List[List[int]]) -> int:
        parent = [i for i in range(n)]
        size = [1]*n

        def findParent(node):
            if parent[node] == node:
                return node
            parent[node] = findParent(parent[node])
            return parent[node]

        def union(first,second):
            firstP = findParent(first)
            secondP = findParent(second)

            if size[firstP]>size[secondP]:
                parent[secondP] = firstP
                size[firstP]+=size[secondP]
            else:
                parent[firstP] = secondP
                size[secondP]+=size[firstP]

        graph= collections.defaultdict(list)
        cycle = []

        for i in edges:
            graph[i[0]].append(i[1])
            graph[i[1]].append(i[0])
            parent1= findParent(i[0])
            parent2 = findParent(i[1])
            if parent1!= parent2:
                union(i[0], i[1])
            else:
                cycle.append(i)
        
        if not cycle:
            return -1

        def getCycleLength(edge):
            visited = set([edge[1]])

            q = [[edge[0],1]]
            ans = 0
            while q:
                cur, count = q.pop(0)
                for nei in graph[cur]:
                    if nei == edge[1] and cur!=edge[0]:
                        return count + 1
                    if nei not in visited:
                        visited.add(nei)
                        q.append([nei, count+1])
        
        ans = math.inf
        for i in cycle:
            ans = min(ans, getCycleLength(i))
        return ans
17400