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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.

Solutions

Solution 1

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class Solution:
    def isBipartite(self, graph: List[List[int]]) -> bool:
        def dfs(u, c):
            color[u] = c
            for v in graph[u]:
                if not color[v]:
                    if not dfs(v, 3 - c):
                        return False
                elif color[v] == c:
                    return False
            return True

        n = len(graph)
        color = [0] * n
        for i in range(n):
            if not color[i] and not dfs(i, 1):
                return False
        return True

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class Solution {
    private int[] color;
    private int[][] g;

    public boolean isBipartite(int[][] graph) {
        int n = graph.length;
        color = new int[n];
        g = graph;
        for (int i = 0; i < n; ++i) {
            if (color[i] == 0 && !dfs(i, 1)) {
                return false;
            }
        }
        return true;
    }

    private boolean dfs(int u, int c) {
        color[u] = c;
        for (int v : g[u]) {
            if (color[v] == 0) {
                if (!dfs(v, 3 - c)) {
                    return false;
                }
            } else if (color[v] == c) {
                return false;
            }
        }
        return true;
    }
}

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class Solution {
public:
    bool isBipartite(vector<vector<int>>& graph) {
        int n = graph.size();
        vector<int> color(n);
        for (int i = 0; i < n; ++i)
            if (!color[i] && !dfs(i, 1, color, graph))
                return false;
        return true;
    }

    bool dfs(int u, int c, vector<int>& color, vector<vector<int>>& g) {
        color[u] = c;
        for (int& v : g[u]) {
            if (!color[v]) {
                if (!dfs(v, 3 - c, color, g)) return false;
            } else if (color[v] == c)
                return false;
        }
        return true;
    }
};

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func isBipartite(graph [][]int) bool {
    n := len(graph)
    color := make([]int, n)
    var dfs func(u, c int) bool
    dfs = func(u, c int) bool {
        color[u] = c
        for _, v := range graph[u] {
            if color[v]