## DEV Community

Abhishek Chaudhary

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# 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 = [[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:
def DFS(self, graph, curr, visited, colors):
for j in graph[curr]:
if j not in visited:
colors[j] = not colors[curr]
if not self.DFS(graph, j, visited, colors):
return False
else:
if colors[j] == colors[curr]:
return False
return True

def isBipartite(self, graph: List[List[int]]) -> bool:
n = len(graph)
for i in range(n):
colors = [False for node in graph]
visited = {i}
if not self.DFS(graph, i, visited, colors):
return False
return True
``````