<p>We want to split a group of <code>n</code> people (labeled from <code>1</code> to <code>n</code>) into two groups of <strong>any size</strong>. Each person may dislike some other people, and they should not go into the same group.</p>

<p>Given the integer <code>n</code> and the array <code>dislikes</code> where <code>dislikes[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that the person labeled <code>a<sub>i</sub></code> does not like the person labeled <code>b<sub>i</sub></code>, return <code>true</code> <em>if it is possible to split everyone into two groups in this way</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 4, dislikes = [[1,2],[1,3],[2,4]]
<strong>Output:</strong> true
<strong>Explanation:</strong> The first group has [1,4], and the second group has [2,3].
</pre>

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

<pre>
<strong>Input:</strong> n = 3, dislikes = [[1,2],[1,3],[2,3]]
<strong>Output:</strong> false
<strong>Explanation:</strong> We need at least 3 groups to divide them. We cannot put them in two groups.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 2000</code></li>
	<li><code>0 &lt;= dislikes.length &lt;= 10<sup>4</sup></code></li>
	<li><code>dislikes[i].length == 2</code></li>
	<li><code>1 &lt;= a<sub>i</sub> &lt; b<sub>i</sub> &lt;= n</code></li>
	<li>All the pairs of <code>dislikes</code> are <strong>unique</strong>.</li>
</ul>

class Solution:
    def possibleBipartition(self, n: int, dislikes: List[List[int]]) -> bool:
        graph = defaultdict(set)

        for u, v in dislikes:
            graph[u].add(v)
            graph[v].add(u)
        
        get_color = defaultdict(lambda: None)
        print(graph)
        self.visited = set()

        def dfs(node, color):    
            get_color[node] = color
            self.visited.add(node)
            
            next_color = "b" if color == "g" else "g"
            for ngb in graph[node]:
                if get_color[ngb] and get_color[ngb] != next_color:
                    return False
                if ngb not in self.visited:
                    if not dfs(ngb, next_color):
                        return False
            return True
        
        for n in graph.keys():
            if n not in self.visited:
                if dfs(n, 'b') == False:
                    return False
        return True

# color: 1 = b, 2 = , 3 = , 4 = 
# 1 - 2
# |   |
# 3   4
24600