<p>You are given a <strong>directed</strong> graph of <code>n</code> nodes numbered from <code>0</code> to <code>n - 1</code>, where each node has <strong>at most one</strong> outgoing edge.</p>

<p>The graph is represented with a given <strong>0-indexed</strong> array <code>edges</code> of size <code>n</code>, indicating that there is a directed edge from node <code>i</code> to node <code>edges[i]</code>. If there is no outgoing edge from node <code>i</code>, then <code>edges[i] == -1</code>.</p>

<p>Return <em>the length of the <strong>longest</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 <strong>same</strong> node.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/06/08/graph4drawio-5.png" style="width: 335px; height: 191px;" />
<pre>
<strong>Input:</strong> edges = [3,3,4,2,3]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The longest cycle in the graph is the cycle: 2 -&gt; 4 -&gt; 3 -&gt; 2.
The length of this cycle is 3, so 3 is returned.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/06/07/graph4drawio-1.png" style="width: 171px; height: 161px;" />
<pre>
<strong>Input:</strong> edges = [2,-1,3,1]
<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>n == edges.length</code></li>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>-1 &lt;= edges[i] &lt; n</code></li>
	<li><code>edges[i] != i</code></li>
</ul>

class Solution:
    def __init__(self):
        self.result = -1  # To store the length of the longest cycle

    def dfs(self, u, edges, visited, count, inRecursion):
        if u == -1 or visited[u]:
            return

        visited[u] = True  # Mark the node as visited
        inRecursion[u] = True  # Mark the node as being part of the current recursion stack
        v = edges[u]  # Get the next node

        if v != -1:
            if not visited[v]:
                count[v] = count[u] + 1  # Update the count for the next node
                self.dfs(v, edges, visited, count, inRecursion)
            elif inRecursion[v]:  # If we encounter a node in the current recursion stack, a cycle is detected
                self.result = max(self.result, count[u] - count[v] + 1)

        inRecursion[u] = False  # Remove the node from the recursion stack

    def longestCycle(self, edges):
        n = len(edges)
        visited = [False] * n  # To track if a node is visited
        count = [1] * n  # To track the length of the path
        inRecursion = [False] * n  # To detect if the node is part of the current recursion stack

        for i in range(n):
            if not visited[i]:
                self.dfs(i, edges, visited, count, inRecursion)

        return self.result

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