# 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: