<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>

#Inputs: edges = List[int]  meaning edge from node i to node edges[i]
#If edges[i] == -1 then there is no edge out of i
#output = node reachable from node1 and node2 with min distance between them
#this node can be one of node1 and node2

#BFS is friend 
#each node has a max of one edge

from collections import deque
class Solution:
    def closestMeetingNode(self, edges: List[int], node1: int, node2: int) -> int:
        #initialize two dicts to keep track of visited nodes for each root node
        #key = nodeVal, val = distFromRoot
        oneDic = {node1: 0}
        twoDic = {node2: 0}
        #initialize minDist to inf
        minDist = math.inf
        minNode = -1

        #Add (node1, 0, 1), (node2, 0, 2) to BFS queue
        bfs = deque([[node1, 0, 1], [node2, 0, 2]])

        #While bfs queue non empty and dist <= minDist
        while bfs and bfs[0][1] <= minDist:
            #pop from queue and check if exists in other sides dict
            node, dist, group = bfs.popleft()

            if group == 1:
                oneDic[node] = dist
                otherDic = twoDic
                dic = oneDic
            else:
                twoDic[node] = dist
                otherDic = oneDic
                dic = twoDic
            
            #if yes, then check dist and replace minDist if necessary
            if node in otherDic:
                maxDist = max(dist, otherDic[node])
                if maxDist < minDist:
                    minDist = maxDist
                    minNode = node
                elif maxDist == minDist:
                    minNode = min(node, minNode)

            #if currNode has an edge then add to bfs
            if edges[node] != -1 and edges[node] not in dic:
                bfs.append([edges[node], dist+1, group])
                
        return minNode
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