<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 kosaraju(self,edges):
        def dfs(hm):
            n = len(hm)
            rtn= []
            seen = [False]*n
            last = [0]*n
            for i in range(n):
                if not seen[i]:
                    stk = [i]
                    seen[i] = True 
                    while stk:
                        curr = stk[-1]
                        for j in range(last[curr], len(hm[curr])):
                            adj = hm[curr][j]
                            if not seen[adj]:
                                seen[adj]= True 
                                last[curr]+=1 
                                stk.append(adj)
                                #break     
                        if curr==stk[-1]: rtn.append(stk.pop())
            return rtn 
        def odfs(curr, seen,hm):
            stk = [curr]
            rtn = 0 
            while stk:
                rtn+=1
                curr = stk.pop()
                for adj in hm[curr]:
                    if not seen[adj]:
                        seen[adj]=True
                        stk.append(adj)
            return rtn 
            
        n = len(edges)
        hm = [[] for _ in range(n)]
        revhm = [[] for _ in range(n)]
        for i,e in enumerate(edges):
            if e!=-1:  
                hm[i].append(e)
                revhm[e].append(i)
        seen = [False]*n
        stk = dfs(hm)
        rtn = 0
        while stk:
            curr = stk.pop()
            if not seen[curr]:
                seen[curr]=True
                rtn = max(odfs(curr,seen,revhm), rtn)
        return rtn if rtn!=1 else -1
        
        

    def longestCycle(self, edges: List[int]) -> int:
        return self.kosaraju(edges)
        
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