'''
785. Is Graph Bipartite?
Medium

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Given an undirected graph, return true if and only if it is bipartite.

Recall that a graph is bipartite if we can split it's set of nodes into two independent subsets A and B such that every edge in the graph has one node in A and another node in B.

The graph is given in the following form: graph[i] is a list of indexes j for which the edge between nodes i and j exists.  Each node is an integer between 0 and graph.length - 1.  There are no self edges or parallel edges: graph[i] does not contain i, and it doesn't contain any element twice.

Example 1:
Input: [[1,3], [0,2], [1,3], [0,2]]
Output: true
Explanation: 
The graph looks like this:
0----1
|    |
|    |
3----2
We can divide the vertices into two groups: {0, 2} and {1, 3}.
Example 2:
Input: [[1,2,3], [0,2], [0,1,3], [0,2]]
Output: false
Explanation: 
The graph looks like this:
0----1
| \  |
|  \ |
3----2
We cannot find a way to divide the set of nodes into two independent subsets.
 

Note:

graph will have length in range [1, 100].
graph[i] will contain integers in range [0, graph.length - 1].
graph[i] will not contain i or duplicate values.
The graph is undirected: if any element j is in graph[i], then i will be in graph[j].
Accepted
92,903
Submissions
200,765
'''

from collections import deque, namedtuple


NodeWColor = namedtuple('NodeWColor',('node', 'color'))


class Solution:

    
    def isBipartite(self, graph: List[List[int]]) -> bool:
        
        if not graph or len(graph) == 0:
            return False
        
    
        
        
        self.visited = {}
        
        for node in range(len(graph)):
            if node not in self.visited and not self.bfs(node, graph):
                return False
        
        return True
            
        
    def bfs(self, node, graph):
        bfs_queue = deque()
        bfs_queue.append(NodeWColor(node=node, color=True))
    
        while len(bfs_queue) > 0:
            
            nodeWColor: NodeWColor = bfs_queue.popleft()
            
            self.visited[nodeWColor.node] = nodeWColor
            
            for neighbor_node in graph[nodeWColor.node]:
                
                if neighbor_node not in self.visited:
                    bfs_queue.append(NodeWColor(node=neighbor_node, color=not nodeWColor.color))
                elif self.visited[neighbor_node].color == nodeWColor.color:
                    return False
                    
        return True