785-is-graph-bipartite¶

Description¶

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(Python)¶

class Solution:
    def isBipartite(self, graph: List[List[int]]) -> bool:
        return self.dfs(graph)

    # Time Complexity: O(V+E)
    # Space Complexity: O(V+E)
    def bfs(self, graph: List[List[int]]) -> bool:
        n = len(graph)
        color = [-1] * n
        q = []

        for i in range(n):
            if color[i] == -1:
                q.append((i, 0))
                color[i] = 0

                while q:
                    u, c = q.pop()

                    for v in graph[u]:
                        if color[v] == c:
                            return False

                        if color[v] == -1:
                            color[v] = 1 - c
                            q.append((v, color[v]))
        return True

    # Time Complexity: O(V+E)
    # Space Complexity: O(V+E)
    def dfs(self, graph: List[List[int]]) -> bool:
        n = len(graph)
        color = {}

        def recur(u, c):
            for v in graph[u]:
                if v not in color:
                    color[v] = 1 - c
                    if not recur(v, color[v]):
                        return False
                if color[v] == c:
                    return False

            return True

        for u in range(n):
            if u not in color:
                color[u] = 0
                if not recur(u, 0):
                    return False
        return True