# 785. Is Graph Bipartite?

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

There are no self-edges (graph[u] does not contain u).

There are no parallel edges (graph[u] does not contain duplicate values).

If v is in graph[u], then u is in graph[v] (the graph is undirected).

The graph may not be connected, meaning there may be two nodes u and v such that there is no path between them.

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

Return true if and only if it is bipartite.

Example 1:

Input: graph = [[1,2,3],[0,2],[0,1,3],[0,2]]
Output: false
Explanation: 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.

Example 2:

Input: graph = [s[1,3],[0,2],[1,3],[0,2]]
Output: true
Explanation: We can partition the nodes into two sets: {0, 2} and {1, 3}.

# Solution

Approach 1: dfs on each node (since garph may not be connected), color them to either 0 or 1, where neighbors can't have the same color.

# Code (Python)

Approach 1:

class Solution:
    def isBipartite(self, graph: List[List[int]]) -> bool:
        # idea: dfs on each node (since garph may not be connected), color them to either 0 or 1, where neighbors can't have the same color.
        color = {}

        def search(node):
            for neighbor in graph[node]:
                if neighbor in color:
                    if color[neighbor] == color[node]:
                        return False
                    # otherwise ok
                else:
                    color[neighbor] = 1 - color[node]
                    if not search(neighbor):
                        return False
            return True
        
        for node in range(len(graph)):
            if node in color:
                continue
            # new unconnected component
            color[node] = 0
            if not search(node):
                return False
        
        return True

# Code (C++)

Approach 1:

Approach 2: