<p>There is a <strong>directed</strong> graph consisting of <code>n</code> nodes numbered from <code>0</code> to <code>n - 1</code> and <code>n</code> directed edges.</p>

<p>You are given a <strong>0-indexed</strong> array <code>edges</code> where <code>edges[i]</code> indicates that there is an edge from node <code>i</code> to node <code>edges[i]</code>.</p>

<p>Consider the following process on the graph:</p>

<ul>
	<li>You start from a node <code>x</code> and keep visiting other nodes through edges until you reach a node that you have already visited before on this <strong>same</strong> process.</li>
</ul>

<p>Return <em>an array </em><code>answer</code><em> where </em><code>answer[i]</code><em> is the number of <strong>different</strong> nodes that you will visit if you perform the process starting from node </em><code>i</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/08/31/graaphdrawio-1.png" />
<pre>
<strong>Input:</strong> edges = [1,2,0,0]
<strong>Output:</strong> [3,3,3,4]
<strong>Explanation:</strong> We perform the process starting from each node in the following way:
- Starting from node 0, we visit the nodes 0 -&gt; 1 -&gt; 2 -&gt; 0. The number of different nodes we visit is 3.
- Starting from node 1, we visit the nodes 1 -&gt; 2 -&gt; 0 -&gt; 1. The number of different nodes we visit is 3.
- Starting from node 2, we visit the nodes 2 -&gt; 0 -&gt; 1 -&gt; 2. The number of different nodes we visit is 3.
- Starting from node 3, we visit the nodes 3 -&gt; 0 -&gt; 1 -&gt; 2 -&gt; 0. The number of different nodes we visit is 4.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/08/31/graaph2drawio.png" style="width: 191px; height: 251px;" />
<pre>
<strong>Input:</strong> edges = [1,2,3,4,0]
<strong>Output:</strong> [5,5,5,5,5]
<strong>Explanation:</strong> Starting from any node we can visit every node in the graph in the process.
</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>0 &lt;= edges[i] &lt;= n - 1</code></li>
	<li><code>edges[i] != i</code></li>
</ul>

class Solution:
    def countVisitedNodes(self, edges: List[int]) -> List[int]:
        
        n = len(edges)
        inorder = [0]*n
        res = [0]*n

        for i,v in enumerate(edges):
            inorder[v] += 1
        
        q = collections.deque()
        for i in range(n):
            if inorder[i] == 0:
                q.append(i)

        non_cycle = set()
        while q:
            node = q.popleft()
            non_cycle.add(node)
            inorder[edges[node]] -= 1
            if inorder[edges[node]] == 0:
                q.append(edges[node])

        visit = set()

        @lru_cache(None)
        def dfs(node):
            visit.add(node)
            if edges[node] not in visit:
                res[node] = 1+dfs(edges[node])
            else:
                res[node] = 1
            return res[node]

        for i in range(n):
            if i not in non_cycle and res[i] == 0:
                cycle_len = dfs(i)
                for node in visit:
                    res[node] = cycle_len
                visit.clear()

        @lru_cache(None)
        def get_len(node):
            if res[node] == 0:
                res[node] = 1+get_len(edges[node])
            return res[node]

        print(res)

        for node in non_cycle:
            get_len(node)

        return res
1489