<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) {
        vector<vector<int>> adj(edges.size(),vector<int>());
        vector<bool> visited(edges.size(),false);
        unordered_map<int,int> inDegree;
        for(int i = 0; i < edges.size(); i++){
            inDegree[i] = 0;
        }
        for(int i = 0; i < edges.size(); i++){
            auto v = edges[i];
            if(v == -1){
                continue;
            }
            auto u = i;
            adj[u].push_back(v); // u -> v
            inDegree[v]++;
        }
        //remove plucking leaf
        deque<int> dq;
        for( auto nd : inDegree){
            if(nd.second == 0){
                dq.push_back(nd.first);
            }
        }
        
       while(!dq.empty()){
            auto n = dq.front();
            dq.pop_front();
            visited[n] = true;
            for(auto neighbour : adj[n]){
                inDegree[neighbour]--;
                if(inDegree[neighbour] == 0){
                    dq.push_back(neighbour);
                }
            }
        }
        
        int answer = -1;
        for (int i = 0; i < edges.size(); i++) {
            if (!visited[i]) {
                int neighbor = edges[i];
                int count = 1;
                visited[i] = true;
                // Iterate in the cycle.
                while (neighbor != i) {
                    visited[neighbor] = true;
                    count++;
                    neighbor = edges[neighbor];
                }
                answer = max(answer, count);
            }
        }
    
        return answer;
    }
};
203425