<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 {
private:
    int cycleLength(int node, int len, int comp, vector<int>& adjlist, vector<int>&visited, vector<int> &path) {

        //already visited node - can be a cycle or part of previous DFS.
        if (visited[node]) {
            if(visited[node]==comp) {   // cycle complete! return the path length
                int i;
                for(i = 0; i < path.size(); i++) if (path[i] == node) break;
                return path.size() - i;
            }     
            return -1;                              // already visited node
        }

        path.push_back(node);

        visited[node] = comp;
        int l = -1;
        if(adjlist[node]!=-1) l = cycleLength(adjlist[node], len + 1, comp, adjlist, visited, path);

        path.pop_back();
        return l;
    }

public:
    int longestCycle(vector<int>& edges) {
        vector<int> visited(edges.size(), 0);
        vector<int> path;

        int component=1;
        int maxLen=-1;
        int len = -1;
        for(int i = 0; i < edges.size(); i++) {
            if(!visited[i])  {
                len = cycleLength(i, 0, component++, edges, visited, path);
                if(maxLen < len){
                    maxLen = len;
                }

            }
        }
        
        return maxLen;
    }
};
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