<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 dfs(vector<vector<int>>&adj, int u, vector<bool>&vis){
        if (vis[u]) return 0;
        vis[u] = true;
        int res = 0;
        for (auto &v : adj[u]){
            if (!vis[v]){
                int prob = dfs(adj, v, vis);
                res = max(res, prob);
            }
        }
        return 1+res;
    }
    int longestCycle(vector<int>& edges) {
        int n = edges.size(); int res = 0; 
        vector<vector<int>>adj(n); 
        vector<int>indeg(n,0); 
        vector<bool>vis(n,false);
        for (int i = 0 ; i < n ; i++){
            int u = i; int v = edges[i];
            if (v!=-1){
                adj[u].push_back(v);
                indeg[v]++;
            }
        }
        //Apply TopoSort to filter the nodes
        queue<int>q;
        for (int i = 0 ; i < n ; i++){
            if (indeg[i] == 0){
                q.push(i);
                vis[i] = 1;
            }
        }
        while(!q.empty()){
            int size = q.size();
            while(size--){
                int node = q.front();
                q.pop();
                for(auto &v : adj[node]){
                    if (indeg[v] > 0){
                        indeg[v]--;
                        if (indeg[v] == 0){
                            q.push(v);
                            vis[v] = true;
                        }
                    }
                }
            }
        }
        
        //Apply BFS and DFS to get longest
        vector<vector<int>>adjT(n);
        for (int i = 0 ; i < n ; i++){
            if (!vis[i]){
                adjT[edges[i]].push_back(i);
            }
        }

        for (int i = 0 ; i < n ; i++){
            if (!vis[i]){
                int prob = dfs(adjT, i, vis);
                res = max(res, prob);
            }
        }
        
        if (res == 0) return -1;
        return res;
    }
};
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