785. Is Graph Bipartite?

Medium (Trung bình) Python 🔗 Xem trên LeetCode

📋 Đề Bài

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.

🧠 Thuật Toán & Kỹ Thuật

DFS (Tìm kiếm theo chiều sâu)Graph (Đồ thị)
⏱️ Thời gian O(V+E)
💾 Không gian O(V)

💻 Lời Giải

Python 0785-is-graph-bipartite.py 
class Solution:
    def isBipartite(self, graph: List[List[int]]) -> bool:
        n = len(graph)
        colors = [0 for _ in range(n)]
        
        def dfs(u: int, color: int) -> bool:
            colors[u] = color
            for v in graph[u]:
                if colors[v] == -color:
                    continue
                if colors[v] == color or not dfs(v, -color):
                    return False
            return True
        
        for i in range(n):
            if colors[i] == 0 and not dfs(i, 1):
                return False
        
        return True