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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

  • 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;
    }
}

  • 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;
    }
};

  • 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


  • 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] == 0 {
				if !dfs(v, 3-c) {
					return false
				}
			} else if color[v] == c {
				return false
			}
		}
		return true
	}
	for i := range graph {
		if color[i] == 0 && !dfs(i, 1) {
			return false
		}
	}
	return true
}

  • function isBipartite(graph: number[][]): boolean {
    const n = graph.length;
    let valid = true;
    let colors = new Array(n).fill(0);
    function dfs(idx: number, color: number, graph: number[][]) {
        colors[idx] = color;
        const nextColor = 3 - color;
        for (let j of graph[idx]) {
            if (!colors[j]) {
                dfs(j, nextColor, graph);
                if (!valid) return;
            } else if (colors[j] != nextColor) {
                valid = false;
                return;
            }
        }
    }

    for (let i = 0; i < n && valid; i++) {
        if (!colors[i]) {
            dfs(i, 1, graph);
        }
    }
    return valid;
}


  • impl Solution {
    #[allow(dead_code)]
    pub fn is_bipartite(graph: Vec<Vec<i32>>) -> bool {
        let mut graph = graph;
        let n = graph.len();
        let mut color_vec: Vec<usize> = vec![0; n];
        for i in 0..n {
            if color_vec[i] == 0 && !Self::traverse(i, 1, &mut color_vec, &mut graph) {
                return false;
            }
        }
        true
    }

    #[allow(dead_code)]
    fn traverse(
        v: usize,
        color: usize,
        color_vec: &mut Vec<usize>,
        graph: &mut Vec<Vec<i32>>
    ) -> bool {
        color_vec[v] = color;
        for n in graph[v].clone() {
            if color_vec[n as usize] == 0 {
                // This node hasn't been colored
                if !Self::traverse(n as usize, 3 - color, color_vec, graph) {
                    return false;
                }
            } else if color_vec[n as usize] == color {
                // The color is the same
                return false;
            }
        }
        true
    }
}

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