<p>There is an <strong>undirected</strong> graph with <code>n</code> nodes, where each node is numbered between <code>0</code> and <code>n - 1</code>. You are given a 2D array <code>graph</code>, where <code>graph[u]</code> is an array of nodes that node <code>u</code> is adjacent to. More formally, for each <code>v</code> in <code>graph[u]</code>, there is an undirected edge between node <code>u</code> and node <code>v</code>. The graph has the following properties:</p>

<ul>
	<li>There are no self-edges (<code>graph[u]</code> does not contain <code>u</code>).</li>
	<li>There are no parallel edges (<code>graph[u]</code> does not contain duplicate values).</li>
	<li>If <code>v</code> is in <code>graph[u]</code>, then <code>u</code> is in <code>graph[v]</code> (the graph is undirected).</li>
	<li>The graph may not be connected, meaning there may be two nodes <code>u</code> and <code>v</code> such that there is no path between them.</li>
</ul>

<p>A graph is <strong>bipartite</strong> if the nodes can be partitioned into two independent sets <code>A</code> and <code>B</code> such that <strong>every</strong> edge in the graph connects a node in set <code>A</code> and a node in set <code>B</code>.</p>

<p>Return <code>true</code><em> if and only if it is <strong>bipartite</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/21/bi2.jpg" style="width: 222px; height: 222px;" />
<pre>
<strong>Input:</strong> graph = [[1,2,3],[0,2],[0,1,3],[0,2]]
<strong>Output:</strong> false
<strong>Explanation:</strong> There is no way to partition the nodes into two independent sets such that every edge connects a node in one and a node in the other.</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/21/bi1.jpg" style="width: 222px; height: 222px;" />
<pre>
<strong>Input:</strong> graph = [[1,3],[0,2],[1,3],[0,2]]
<strong>Output:</strong> true
<strong>Explanation:</strong> We can partition the nodes into two sets: {0, 2} and {1, 3}.</pre>

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

<ul>
	<li><code>graph.length == n</code></li>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code>0 &lt;= graph[u].length &lt; n</code></li>
	<li><code>0 &lt;= graph[u][i] &lt;= n - 1</code></li>
	<li><code>graph[u]</code>&nbsp;does not contain&nbsp;<code>u</code>.</li>
	<li>All the values of <code>graph[u]</code> are <strong>unique</strong>.</li>
	<li>If <code>graph[u]</code> contains <code>v</code>, then <code>graph[v]</code> contains <code>u</code>.</li>
</ul>

from collections import deque 
class Solution:
    def isBipartite(self, graph: List[List[int]]) -> bool:
        """
        start from 0 -> n 
        color each item in the neighbor based on the current color 
        """

        N = len(graph)
        # O(V)
        colors = [-1 for _ in range(N)]
        visited = set()

        # O(V + E)
        for node in range(N):

            if node in visited:
                continue
            q = deque([node])
            colors[node] = 0
            while q:
                curNode = q.popleft()
                visited.add(curNode)
                nbColor = 0 if colors[curNode] == 1 else 1
                for nb in graph[curNode]:
                    if colors[nb] == colors[curNode]:
                        return False 
                    colors[nb] = nbColor
                    if nb not in visited:
                        q.append(nb)

        return True 
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