<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 deque
from collections import defaultdict

class Solution:
    def magnificentSets(self, n: int, edges: List[List[int]]) -> int:
        edges_arr = defaultdict(set)
        for edge in edges:
            edges_arr[edge[0]].add(edge[1])
            edges_arr[edge[1]].add(edge[0])

        not_computed = set([i + 1 for i in range(n)])
        max_group = {}

        while not_computed:
            curr = not_computed.pop()
            groups, visited = bfs(edges_arr, curr, n)
            if groups == -1:
                return -1
            str_v = str(visited)
            if str_v not in max_group:
                max_group[str_v] = -2
            max_group[str_v] = max(groups, max_group[str_v])
        return sum(max_group.values())

# return number of groups. start is 1
def bfs(edges, start, n):
    # visited also 1 indexed
    visited = [0]*(n+1)
    visited[start] = 1
    
    color = [0]*(n+1)
    color[start] = 1
    # q is also 0 indexed
    q = deque([start])

    depth = 0
    curr_color = 1
    while q:
        depth += 1
        curr_color = set_color(curr_color)
        for _ in range(len(q)): # this does not add runtime but it splits the bfs into layers
            # curr is also 1 indexed
            curr = q.popleft()
            for node in edges[curr]:
                    if not valid_color(color[curr], color[node]):
                        return -1, visited
                    if visited[node] == 0:
                        q.append(node)
                        visited[node] = 1
                        color[node] = set_color(color[curr])
    return depth, visited

def valid_color(a, b):
    if a == 1 and b == 1:
        return False
    if a == 2 and b == 2:
        return False
    return True

def set_color(a):
    if a == 1:
        return 2
    if a == 2:
        return 1
    return 0
21000