# Is Graph Bipartite?

Updated: Mar 25, 2021

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

**Constraints:**

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.

**Solution:**

```
class Solution {
public boolean isBipartite(int[][] graph) {
int[] colors = new int[graph.length];
Queue<Integer> queue = new LinkedList<>();
for(int i=0;i<graph.length;i++)
{
if(colors[i]==0)
{
colors[i] = 1;
queue.add(i);
while(!queue.isEmpty())
{
int j = queue.poll();
for(int neighbors:graph[j])
{
if(colors[neighbors]==0)
{
colors[neighbors]= -colors[j];
queue.add(neighbors);
}
if(colors[j]==colors[neighbors]) return false;
}
}
}
}
return true;
}
}
```