<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>

from collections import deque

class Solution:
    def countVisitedNodes(self, edges: List[int]) -> List[int]:
        visited = [0 for _ in edges]
        computed_nodes = set()
        node_count = len(visited)
        for n in range(node_count):
            current = n
            if current in computed_nodes:
                continue
            seen = set()
            traversal: Deque[int] = deque()
            # keep traversing until we encouter a duplicate OR
            # we reach a node that has already been computed
            while current not in seen and current not in computed_nodes:
                traversal.append(current)
                seen.add(current)
                current = edges[current]
            visited_count = len(traversal)

            if current in computed_nodes:
                # We have reached a pre-computed node, so
                # we can backtrack to fill in the nodes we have visited
                # until this point
                visit_count = visited[current]
                while traversal:
                    visit_count = visit_count + 1
                    current = traversal.pop()
                    visited[current] = visit_count
                    computed_nodes.add(current)
                continue

            # Else we encountered a cycle
            cycle_node = current
            # Mark any nodes that are not a part of the cycle
            # Each node we skip from start, we are essentially
            # trunctating the cycle from that node, until we reach
            # the cycle node
            while traversal and traversal[0] != cycle_node:
                node = traversal.popleft()
                computed_nodes.add(node)
                visited[node] = visited_count
                visited_count = visited_count - 1
            # Now we have reached the cycle
            while traversal:
                node = traversal.popleft()
                computed_nodes.add(node)
                visited[node] = visited_count
        return visited
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