<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:
        edge_dict = defaultdict(set)
        for f, t in edges:
            edge_dict[f].add(t)
            edge_dict[t].add(f)

        self.shortest_cycle = float("inf")
        def bfs(node):
            visited = [False for _ in range(n)]
            dis = [-1 for _ in range(n)]
            queue = deque([(node, node)])
            dis[node] = 0
            visited[node] = True
            while queue:
                node, prev = queue.popleft()
                neighbors = edge_dict[node]
                for neigh in neighbors.copy():
                    if neigh == prev:
                        continue
                    if visited[neigh]:
                        self.shortest_cycle = min(self.shortest_cycle, dis[neigh] + dis[node] + 1)
                        continue
                    queue.append((neigh, node))
                    dis[neigh] = dis[node] + 1
                    visited[neigh] = True

        for node in range(n):
            bfs(node)

        if self.shortest_cycle == float("inf"):
            return -1
        return self.shortest_cycle
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