<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>

from collections import deque
from typing import List

class Solution:
    def findShortestCycle(self, n: int, edges: List[List[int]]) -> int:
        # Create adjacency list for the graph
        graph = [[] for _ in range(n)]
        for u, v in edges:
            graph[u].append(v)
            graph[v].append(u)
        
        def bfs(start):
            # Initialize distances and parent tracking
            dist = [-1] * n
            dist[start] = 0
            queue = deque([(start, -1)])  # (current node, parent node)
            min_cycle_length = float('inf')
            
            while queue:
                node, parent = queue.popleft()
                
                for neighbor in graph[node]:
                    if dist[neighbor] == -1:  # If not visited
                        dist[neighbor] = dist[node] + 1
                        queue.append((neighbor, node))
                    elif neighbor != parent:  # Found a cycle
                        min_cycle_length = min(min_cycle_length, dist[node] + dist[neighbor] + 1)
            
            return min_cycle_length
        
        # Find the shortest cycle
        shortest_cycle = float('inf')
        for i in range(n):
            shortest_cycle = min(shortest_cycle, bfs(i))
        
        return shortest_cycle if shortest_cycle != float('inf') else -1
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