<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>

import collections
class Solution:
  def closestMeetingNode(self, edges: List[int], node1: int, node2: int) -> int:
    '''
    use BFS to keep track of node reached & distance from node 1, node 2
    '''
    if node1 == node2:
      return node1
    
    def dist(node):
      d = {node: 0}
      while node != -1 and edges[node] not in d:
        next_node = edges[node]
        d[next_node] = d[node] + 1
        node = next_node
      return d
    d1, d2 = dist(node1), dist(node2)
    common = ((max(d1[node], d2[node]), node) for node in d1.keys() & d2.keys())
    return min(common, default=(0, -1))[1]
    # visited1, visited2 = [], []
    # n = len(edges)
    # common = []
    # while n > 0:
    #   if node1 == node2:
    #     common.append(node1)
    #     break
    #   if node1 in visited2 or node2 in visited1:
    #     if node1 in visited2:
    #       common.append(node1)
    #     if node2 in visited1:
    #       common.append(node2)
    #     break
    #   if node1 != -1:
    #     visited1.append(node1)
    #     node1 = edges[node1]
    #   if node2 != -1:
    #     visited2.append(node2)
    #     node2 = edges[node2]
    #   n -= 1
    # print(common, visited1, visited2)
    # return min(common) if common else -1
    
    # d1 = defaultdict(int)
    # curr = edges[node1]
    # d = 0
    # d1[node1] = 0
    # while not (curr == -1 or curr in d1):
    #   d += 1
    #   d1[curr] = d
    #   curr = edges[curr]
    
    # d2 = defaultdict(int)
    # curr = edges[node2]
    # d = 0
    # d2[node2] = 0
    # while not (curr == -1 or curr in d2):
    #   d += 1
    #   d2[curr] = d
    #   curr = edges[curr]
    # print(d1, d2)
    # keys = d1.keys() & d2.keys()
    # print(keys)
    # if not keys:
    #   return -1
    # min, ans = float('inf'), -1
    # for k in keys:
    #   if (d1[k] + d2[k]) < min:
    #     min = (d1[k] + d2[k])
    #     ans = k
    # return ans 
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