Namaste, I'm Avinash Sharma.

Is Graph Bipartite?

published on 4/29/2022

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:

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


  • graph.length == n
  • 1 <= n <= 100
  • 0 <= graph[u].length < n
  • 0 <= graph[u][i] <= n - 1
  • graph[u] does not contain u.
  • All the values of graph[u] are unique.
  • If graph[u] contains v, then graph[v] contains u.


  • If a graph has an odd length cycle, then it's not a bipartite graph.
  • If it doesn't have an odd length cycle, then it's a bipartite graph.

Solution in DFS:

Solution explanation:

  • Initialize the color array with -1
  • We can use a color array to mark the nodes, wether colored or not.
  • We can use DFS to check if the graph is bipartite.
  • If the node is not colored, then we color its with 1.
  • And run DFS on its neighbors.
  • If the current node is not colored, then assign it with 1-color[node]
  • If the node is colored,and its equal to the previous node color then it's not bipartite.
class Solution:
def isBipartite(self, graph: List[List[int]]) -> bool:
n = len(graph)
color = [-1 for i in range(n)]
def dfs(node):
if color[node] == -1:
color[node] = 1
for curNode in graph[node]:
if color[curNode] == -1:
color[curNode] = 1-color[node]
if(not dfs(curNode)):
return False
elif color[curNode] == color[node]:
return False
return True
for i in range(n):
if color[i] == -1:
if(not dfs(i)):
return False
return True

Time complexity: O(N+E)

Space complexity: O(N+E)+O(N)+O(N)

Created with ❤️ by Avinash Sharma