<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]:

        def find(u):
            if u != par[u]:
                par[u] = find(par[u])
            return par[u]

        def union(u, v):
            x, y = find(u), find(v)
            if x != y: 
                par[x] = y

        par = list(range(len(edges)))
        rank = [0] * len(edges)
        rev = defaultdict(list)
        for u, v in enumerate(edges):
            rev[v].append(u)
            rank[v] += 1
            union(u, v)

        comp = defaultdict(list)
        for i in range(len(edges)):
            comp[find(i)].append(i)

        ans = [0] * len(edges)
        def dfs(n, i):      
            ans[n] = len(seen) + i
            for nei in rev[n]:
                if nei not in seen:
                    dfs(nei, i + 1)

        for v in comp.values():
            tortoise = hare = v[0]
            while True:
                tortoise = edges[tortoise]
                hare = edges[edges[hare]]
                if tortoise == hare:
                    break

            tortoise = v[0]
            while tortoise != hare:
                tortoise = edges[tortoise]
                hare = edges[hare]

            seen = set()
            while hare not in seen:
                seen.add(hare)
                hare = edges[hare]

            for i in seen:
                dfs(i, 0)

        return ans






            
1858