LeetCode 785. Is Graph Bipartite?

Description

https://leetcode.com/problems/design-add-and-search-words-data-structure/

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.

Explanation

We can use the coloring approach with either breadth-first search or depth-first search, to see if nodes can be divided into half.

Python Solution

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

        visited = set()
        colors = {}
        
        for i in range(len(graph)):
            if i in visited:
                continue
                        
            queue = []
            queue.append(i)
            colors[i] = 0
            visited.add(i)
            
            while queue:
                node = queue.pop(0)                
                
                for neighbor in graph[node]:
                    if neighbor not in colors:
                        colors[neighbor] = 1 if colors[node] == 0 else 0
                        queue.append(neighbor)
                        visited.add(neighbor)                        
                    else:
                        if colors[neighbor] == colors[node]:
                            return False
                        
        return True
         
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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