<p>You are given a positive integer <code>n</code> representing the number of nodes in an <strong>undirected</strong> graph. The nodes are labeled from <code>1</code> to <code>n</code>.</p>

<p>You are also given a 2D integer array <code>edges</code>, where <code>edges[i] = [a<sub>i, </sub>b<sub>i</sub>]</code> indicates that there is a <strong>bidirectional</strong> edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code>. <strong>Notice</strong> that the given graph may be disconnected.</p>

<p>Divide the nodes of the graph into <code>m</code> groups (<strong>1-indexed</strong>) such that:</p>

<ul>
	<li>Each node in the graph belongs to exactly one group.</li>
	<li>For every pair of nodes in the graph that are connected by an edge <code>[a<sub>i, </sub>b<sub>i</sub>]</code>, if <code>a<sub>i</sub></code> belongs to the group with index <code>x</code>, and <code>b<sub>i</sub></code> belongs to the group with index <code>y</code>, then <code>|y - x| = 1</code>.</li>
</ul>

<p>Return <em>the maximum number of groups (i.e., maximum </em><code>m</code><em>) into which you can divide the nodes</em>. Return <code>-1</code> <em>if it is impossible to group the nodes with the given conditions</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/10/13/example1.png" style="width: 352px; height: 201px;" />
<pre>
<strong>Input:</strong> n = 6, edges = [[1,2],[1,4],[1,5],[2,6],[2,3],[4,6]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> As shown in the image we:
- Add node 5 to the first group.
- Add node 1 to the second group.
- Add nodes 2 and 4 to the third group.
- Add nodes 3 and 6 to the fourth group.
We can see that every edge is satisfied.
It can be shown that that if we create a fifth group and move any node from the third or fourth group to it, at least on of the edges will not be satisfied.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 3, edges = [[1,2],[2,3],[3,1]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> If we add node 1 to the first group, node 2 to the second group, and node 3 to the third group to satisfy the first two edges, we can see that the third edge will not be satisfied.
It can be shown that no grouping is possible.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 500</code></li>
	<li><code>1 &lt;= edges.length &lt;= 10<sup>4</sup></code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>1 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt;= n</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li>There is at most one edge between any pair of vertices.</li>
</ul>

from collections import defaultdict, deque

class Solution:
    def magnificentSets(self, n: int, edges: List[List[int]]) -> int:
        adj = defaultdict(list)
        for u, v in edges:
            adj[u].append(v)
            adj[v].append(u)

        def is_bipartite(node):
            color[node] = 0
            q = deque([node])
            comp = [node]
            while q:
                u = q.popleft()
                for v in adj[u]:
                    if color[v] == -1:
                        color[v] = 1 - color[u]
                        q.append(v)
                        comp.append(v)
                    elif color[v] == color[u]:
                        return False, []
            return True, comp

        def bfs(node):
            dist = {}
            q = deque([node])
            dist[node] = 1
            while q:
                u = q.popleft()
                for v in adj[u]:
                    if v not in dist:
                        dist[v] = dist[u] + 1
                        q.append(v)
            return max(dist.values())

        color = [-1] * (n + 1)
        res = 0

        for i in range(1, n + 1):
            if color[i] == -1:
                bipartite, comp = is_bipartite(i)
                if not bipartite:
                    return -1
                max_groups = 0
                for node in comp:
                    depth = bfs(node)
                    max_groups = max(max_groups, depth)
                res += max_groups
        return res

20500