<p>You are given a <strong>directed</strong> graph of <code>n</code> nodes numbered from <code>0</code> to <code>n - 1</code>, where each node has <strong>at most one</strong> outgoing edge.</p>

<p>The graph is represented with a given <strong>0-indexed</strong> array <code>edges</code> of size <code>n</code>, indicating that there is a directed edge from node <code>i</code> to node <code>edges[i]</code>. If there is no outgoing edge from <code>i</code>, then <code>edges[i] == -1</code>.</p>

<p>You are also given two integers <code>node1</code> and <code>node2</code>.</p>

<p>Return <em>the <strong>index</strong> of the node that can be reached from both </em><code>node1</code><em> and </em><code>node2</code><em>, such that the <strong>maximum</strong> between the distance from </em><code>node1</code><em> to that node, and from </em><code>node2</code><em> to that node is <strong>minimized</strong></em>. If there are multiple answers, return the node with the <strong>smallest</strong> index, and if no possible answer exists, return <code>-1</code>.</p>

<p>Note that <code>edges</code> may contain cycles.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/06/07/graph4drawio-2.png" style="width: 321px; height: 161px;" />
<pre>
<strong>Input:</strong> edges = [2,2,3,-1], node1 = 0, node2 = 1
<strong>Output:</strong> 2
<strong>Explanation:</strong> The distance from node 0 to node 2 is 1, and the distance from node 1 to node 2 is 1.
The maximum of those two distances is 1. It can be proven that we cannot get a node with a smaller maximum distance than 1, so we return node 2.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/06/07/graph4drawio-4.png" style="width: 195px; height: 161px;" />
<pre>
<strong>Input:</strong> edges = [1,2,-1], node1 = 0, node2 = 2
<strong>Output:</strong> 2
<strong>Explanation:</strong> The distance from node 0 to node 2 is 2, and the distance from node 2 to itself is 0.
The maximum of those two distances is 2. It can be proven that we cannot get a node with a smaller maximum distance than 2, so we return node 2.
</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>-1 &lt;= edges[i] &lt; n</code></li>
	<li><code>edges[i] != i</code></li>
	<li><code>0 &lt;= node1, node2 &lt; n</code></li>
</ul>

class Solution:
    def closestMeetingNode(self, edges: List[int], node1: int, node2: int) -> int:
        """
        Idea: start by doing a DFS from node2.
        As we hit each node, "label it" with the distance from node 2 (we can use another list for this)
        Since each node only has one outgoing edge, we can just have an ordered list of nodes reachable from 2
        We can also have an ordered list of nodes reachable from 1 (ordered by distance)

        We can use two pointers, and record the distances and find the best one based on our iteration
        Issue:
        Let's say that I had node2 edges and node1 edges. I want to maximize distance from node1, so 
        we start at the end of node1's list. The thing is, I would need to check every node in node2's list
        and compare the max and minimum distances.

        So this would end up being O(e^2), but that is fine? (hopefully)

        Let's begin
        """


        def dfs(node: int) -> List[int]:
            cycle = set()
            nodes = [node]
            output = []
            while nodes:
                current_node = nodes.pop()
                if current_node in cycle or current_node == -1:
                    return output
                cycle.add(current_node)
                output.append(current_node)
                nodes.append(edges[current_node])
            return output
        
        nodes_reachable_from_1 = dfs(node1)
        nodes_reachable_from_2 = dfs(node2)
        dist_from_node_1 = {}
        for distance, node in enumerate(nodes_reachable_from_1):
            dist_from_node_1[node] = distance

        min_distance = float("infinity")
        best_node = -1

        for distance_from_2, node in enumerate(nodes_reachable_from_2):
            if node in dist_from_node_1:
                max_distance = max(dist_from_node_1[node], distance_from_2)
                if max_distance < min_distance:
                    min_distance = max_distance
                    best_node = node
                elif max_distance == min_distance:
                    best_node = min(node, best_node)
    
        return best_node



38288