<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 result = -1;  // To store the length of the longest cycle

    void dfs(int u, vector<int>& edges, vector<int>& visited, vector<int>& count, vector<bool>& inRecursion) {
        if (u == -1 || visited[u]) {
            return;
        }

        visited[u] = true;       // Mark the node as visited
        inRecursion[u] = true;   // Mark the node as being part of the current recursion stack
        int v = edges[u];        // Get the next node

        if (v != -1) {
            if (!visited[v]) {
                count[v] = count[u] + 1;  // Update the count for the next node
                dfs(v, edges, visited, count, inRecursion);
            } else if (inRecursion[v]) {  // If we encounter a node that is in the current recursion stack, it means we found a cycle
                result = max(result, count[u] - count[v] + 1);
            }
        }

        inRecursion[u] = false;  // Remove the node from the recursion stack
    }

    int longestCycle(vector<int>& edges) {
        int n = edges.size();
        vector<int> visited(n, 0);   // To track if a node is visited
        vector<int> count(n, 1);     // To track the length of the path
        vector<bool> inRecursion(n, false);  // To detect if the node is part of the current recursion stack

        for (int i = 0; i < n; ++i) {
            if (!visited[i]) {
                dfs(i, edges, visited, count, inRecursion);
            }
        }

        return result;
    }
};

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