<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>

using ll = long long ;
using vl = vector<ll> ;
using vvl = vector<vl> ; 

class Solution {
public: 

    ll active_node = 1;
    ll expired_node = 2; 
    
    void dfs(ll cur,ll &ans,ll depth, vl &vis,vl& dp, vvl &g){

        if(vis[cur]==active_node){
            ans = max(ans,depth-dp[cur]);return ;
        }

        if(vis[cur]==expired_node){
            return ;
        }

        vis[cur] = active_node;
        dp[cur] = min(dp[cur],depth);

        for(auto &&child : g[cur])dfs(child,ans,depth+1ll,vis,dp,g);

        vis[cur] = expired_node;
       
    }

    int longestCycle(vector<int>& ed) {

        ll n = ed.size();vvl g(n);
        for(ll i=0;i<n;i++)if(ed[i]!=-1)g[i].push_back(ed[i]);

        vl dp(n,1e16);vl vis(n,0);ll flag = -1; 
        for(ll i=0;i<n;i++)if(!vis[i])dfs(i,flag,0ll,vis,dp,g);

        return (int)flag==0?-1:flag;

 
    }
};
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