<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>

class Solution:
    def magnificentSets(self, n: int, edges: List[List[int]]) -> int:
    
        edge_map = {i: set() for i in range(1,n+1)}
        connected_comps = []
        total_seen = set()

        for edge in edges:
            edge_map[edge[0]].add(edge[1])
            edge_map[edge[1]].add(edge[0])
        
        def is_graph_bipartite(node, visited, color):
            visited[node] = color
            for nei in edge_map[node]:
                if nei in visited:
                    if visited[nei] == color:
                        return False
                else:
                    if not is_graph_bipartite(nei, visited, ~color):
                        return False
            return True
        
        for i in range(1, n+1):
            if i not in total_seen:
                visited = {}
                if not is_graph_bipartite(i, visited, 0):
                    return -1
                connected_comps.append(visited.keys())
                total_seen.update(visited.keys())
        
        def bfs(node):
            visited = set()
            que = deque([node])
            visited.add(node)
            level = 0
            while len(que):
                level+=1
                for _ in range(len(que)):
                    node = que.popleft()
                    for nei in edge_map[node]:
                        if nei not in visited:
                            que.append(nei)
                            visited.add(nei)
            return level
        
        ans = 0
        for comp in connected_comps:
            comp_max_groups = 0
            for node in comp:
                comp_max_groups = max(comp_max_groups, bfs(node))
            ans+=comp_max_groups
        return ans
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