<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 {
public:
    int longestCycle(vector<int>& edges) {
        set<int> possible;
        int n=edges.size();
        vector<vector<int>> adj(n);
        for(int i=0;i<edges.size();i++){
            if(edges[i]==-1) continue;
            adj[i].push_back(edges[i]);
        }
        map<int,int> parent;
        parent[0]=-1;
        vector<int> visited1(n);
        // auto dfs=[&](auto &&self,int i)->void{
        //     visited1[i]=true;
        //     for(auto it:adj[i]){
        //         if(!visited1[i]){
        //         parent[it]=i;
        //         self(self,it);
        //         }
        //     }

        // };
        bool flag=true;
        vector<int> pathvisited(edges.size());
        vector<int> visited(edges.size());
                auto check=[&](auto &&self,int i)->bool{
                        visited[i]=1;
                        pathvisited[i]=1;
                        for(auto it:adj[i]){
                            if(!visited[it]){
                                if(self(self,it)) {
                                    if(possible.find(it)!=possible.end()) flag=false;
                                    if(flag) possible.insert(it);
                                    return true;
                            }
                            }
                            else if(pathvisited[it]) {
                                possible.insert(it);
                                return true;
                            }
                        }
                        visited[i]=0;
                        pathvisited[i]=0;
                        return false;
                
                
                };
                //if(!check(check,0)) return -1;
                
                int ans=-1;
                for(int i=0;i<edges.size();i++){
                    if(!visited[i]){
                        if(check(check,i)){
                            ans=max(ans,(int)possible.size());
                            flag=true;
                            possible.clear();
                            pathvisited.assign(n,0);
                        }
                    }
                }
                return ans;

        
    
    }
};
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