<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) {
        int ans = -1;
        // topological sort
        vector<int> indegree(edges.size()); 
        unordered_set<int> visit;

        // calculate in-degree
        for (auto e : edges) {
            if (e != -1) indegree[e]++;
        } 

        queue<int> queue;
        
        // queue up the edges with indegree 
        for (int i = 0; i < indegree.size(); ++i) {
            if (indegree[i] == 0) queue.push(i);
        }

        // go through the queue
        while (!queue.empty()) {
            int curr = queue.front();
            visit.insert(curr);
            queue.pop();

            // traverse through the edges using node as the index
            int neighbor = edges[curr];
            if (neighbor != -1) {
                indegree[neighbor]--;
                // add to queue if indegree has became 0
                if (indegree[neighbor] == 0) queue.push(neighbor);
            }
        }

        // We should have traversed through nodes that have had in-degree 0 
        // Now we traverse again through all edges check the indegree and loop again to see if there's a cycle present
        for (int i = 0; i < edges.size(); ++i) {
            if (!visit.count(i)) {
                int count = 1;
                int curr = edges[i]; 
                while (curr != i) {
                    ++count;
                    visit.insert(curr);
                    // traverse through list
                    curr = edges[curr];
                }
                // we want the maximum so we compare the current count to the ans 
                ans = max(ans, count);
            }
        } 

        return ans;
    }
};
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