<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]:
        deg = defaultdict(int)
        for u in edges : deg[u] += 1
        n = len(edges)
        
        # 1. FIND CYCLIC NODES
        
        # starting with nodes without in-edges
        Q = deque(u for u in range(n) if deg[u] == 0)
        
        # proceed until we're left with cyclic nodes
        while Q:
            for _ in range(len(Q)):
                u = Q.popleft()
                v = edges[u]
                deg[v] -= 1
                if deg[v] == 0:
                    Q.append(v)
        
        # 2. PROCESS CYCLES
        
        count = defaultdict(int)
        vis = set()
        
        # try starting cycle from each of the nodes
        # that were left after the previous step
        for u in filter(lambda u: deg[u] != 0, range(n)):
            if u in vis : continue
            
            # build the cycle
            cycle = []
            while u not in vis:
                cycle.append(u)
                vis.add(u)
                u = edges[u]
            
            # this is the answer for each of the cycle nodes
            for v in cycle : count[v] = len(cycle)

        # 3. PROCESS NON-CYCLIC NODES
        
        # increse path length by 1 or return the cycle 
        # length when encountered one of the cycle nodes
        @cache
        def dfs(v):
            if v in count : return count[v]
            return 1 + dfs(edges[v])
        
        # all results are now @cached, we have to collect them in order
        return [dfs(v) for v in range(n)]
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