<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:
    
    // dfs function for finding the length of cycle
    
    void dfs(vector<int>& edges, int u, vector<bool>& vis, int &count)
    {
        vis[u] = true;
        
        count++;
        
        if(edges[u] != -1 && vis[edges[u]] == false)
        {
            dfs(edges, edges[u], vis, count);
        }
    }
    
    int longestCycle(vector<int>& edges) {
        
        int n = edges.size();
        
        // use kahn's algorithm
        
        vector<int> indegree(n, 0);
        
        for(int i = 0; i < n; i++)
        {
            int v = edges[i];
            
            if(v != -1)
            {
                indegree[v]++;
            }
        }
        
        queue<int> q;
        
        for(int i = 0; i < n; i++)
        {
            if(indegree[i] == 0)
            {
                q.push(i);
            }
        }
        
        while(!q.empty())
        {
            int u = q.front();
            
            q.pop();
            
            if(edges[u] != -1)
            {
                indegree[edges[u]]--;
                
                if(indegree[edges[u]] == 0)
                {
                    q.push(edges[u]);
                }
            }
        }
        
        // declare a vis array
        
        vector<bool> vis(n, false);
        
        int maxi = INT_MIN;
        
        // run a dfs for every node with indegree > 0
        
        // find the length of cycle and update maxi
        
        for(int i = 0; i < n; i++)
        {
            int count = 0;
            
            if(indegree[i] > 0 && vis[i] == false)
            {
                dfs(edges, i, vis, count);
            }
            
            maxi = max(maxi, count);
        }
        
        // if there is no cycle in graph, return -1
        
        if(maxi == 0)
            return -1;
        
        return maxi;
    }
};
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