Is Graph Bipartite?

#leetcode #dsa #theabbie

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:
                visited.add(j)
                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