<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 V = edges.size();
        vector<int> indegree(V, 0);  // Store in-degree of each node

        // Step 1: Build the graph and in-degree array
        for (int i = 0; i < V; ++i)
        {
            if (edges[i] != -1) // Only count valid edges
                indegree[edges[i]]++;
        }

        // Step 2: Topological Sort using BFS (Kahn's Algorithm)
        queue<int> q;
        vector<bool> visited(V, false);

        for (int i = 0; i < V; ++i)
        {
            if (indegree[i] == 0) // Start with nodes having zero in-degree
                q.push(i);
        }

        // Remove all nodes that are part of acyclic paths
        while (!q.empty())
        {
            int node = q.front();
            q.pop();
            visited[node] = true;

            int neighbor = edges[node];
            if (neighbor != -1)
            {
                indegree[neighbor]--;
                if (indegree[neighbor] == 0)  // If indegree becomes 0, add to the queue
                    q.push(neighbor);
            }
        }

        // Step 3: Find the longest cycle
        int longestCycleLength = -1;

        for (int i = 0; i < V; ++i)
        {
            if (!visited[i])  // Unvisited nodes are part of cycles
            {
                int cycleLength = 0;
                int start = i;

                // Traverse the cycle
                while (!visited[start])
                {
                    visited[start] = true;
                    start = edges[start];
                    cycleLength++;
                }

                longestCycleLength = max(longestCycleLength, cycleLength);
                  // Keep track of the longest cycle
            }
        }

        return longestCycleLength;
    }
};
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