<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:
        # BFS function to calculate distances from a starting node
        def bfs(start, dist_map):
            q = collections.deque([start])
            dist_map[start] = 0
            dist = 1
            while q:
                node = q.popleft()
                nei = edges[node]  # Follow the edge
                if nei != -1 and nei not in dist_map:  # Check if the neighbor is valid and not visited
                    dist_map[nei] = dist
                    q.append(nei)
                dist += 1

        # Distance maps for node1 and node2
        dist_from1 = {}
        dist_from2 = {}

        # Run BFS from both nodes
        bfs(node1, dist_from1)
        bfs(node2, dist_from2)

        # Result variables
        res = -1
        min_dist = float('inf')

        # Iterate over all nodes to find the closest meeting node
        for i in range(len(edges)):
            if i not in dist_from1 or i not in dist_from2:
                continue
            max_dist = max(dist_from1[i], dist_from2[i])  # Choose the max of the two distances
            if max_dist < min_dist:
                min_dist = max_dist
                res = i
            elif max_dist == min_dist:  # Tie-breaking for smaller index
                res = min(res, i)

        return res



    def closestMeetingNode2(self, edges: List[int], node1: int, node2: int) -> int:
        graph = {i: [] for i in range(len(edges))}

        for i in range(len(edges)):
            if edges[i] == -1:
                continue
            graph[i].append(edges[i])

        def bfs(i, dist_map):
            q = collections.deque()
            q.append(i)
            dist_map[i] = 0
            dist = 1
            while q:
                for _ in range(len(q)):
                    node = q.popleft()
                    for nei in graph[node]:
                        if nei in dist_map:
                            continue
                        dist_map[nei] = dist
                        q.append(nei)
                dist += 1

        dist_from1 = {}
        dist_from2 = {}
        bfs(node1, dist_from1)
        bfs(node2, dist_from2)

        res = -1
        min_dist = float(inf)
        for i in range(len(edges)):
            if i not in dist_from1 or i not in dist_from2:
                continue
            dist = max(dist_from1[i], dist_from2[i])
            if dist < min_dist:
                res = i
                min_dist = dist

        return res         


    # what if equality?
    def closestMeetingNode_not(self, edges: List[int], node1: int, node2: int) -> int:
        graph = {i: [] for i in range(len(edges))}

        for i in range(len(edges)):
            if edges[i] == -1:
                continue
            graph[i].append(edges[i])

        q = collections.deque()
        q.append((-1, node1))
        q.append((-1, node2))

        visited = set()

        while q:
            prev, node = q.popleft()
            if node in visited:
                return node
            visited.add(node)
            for nei in graph[node]:
                if nei == prev:
                    continue
                q.append((node, nei))

        return -1
853