Count Visited Nodes in a Directed Graph
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2800-2899/2876.Count%20Visited%20Nodes%20in%20a%20Directed%20Graph/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2800-2899/2876.Count%20Visited%20Nodes%20in%20a%20Directed%20Graph/images/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>
Graph; Memoization; Dynamic Programming
class Solution {
public:
    vector<int> countVisitedNodes(vector<int>& edges) {
        int n = edges.size();
        vector<int> ans(n), vis(n);
        for (int i = 0; i < n; ++i) {
            if (!ans[i]) {
                int cnt = 0, j = i;
                while (vis[j] == 0) {
                    vis[j] = ++cnt;
                    j = edges[j];
                }
                int cycle = 0, total = cnt + ans[j];
                if (ans[j] == 0) {
                    cycle = cnt - vis[j] + 1;
                }
                j = i;
                while (ans[j] == 0) {
                    ans[j] = max(total--, cycle);
                    j = edges[j];
                }
            }
        }
        return ans;
    }
};