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 {
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
}

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
}
}