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Formatted question description: https://leetcode.ca/all/785.html

# 785. Is Graph Bipartite? (Medium)

Given an undirected graph, return true if and only if it is bipartite.

Recall that a graph is bipartite if we can split it's set of nodes into two independent subsets A and B such that every edge in the graph has one node in A and another node in B.

The graph is given in the following form: graph[i] is a list of indexes j for which the edge between nodes i and j exists.  Each node is an integer between 0 and graph.length - 1.  There are no self edges or parallel edges: graph[i] does not contain i, and it doesn't contain any element twice.

Example 1:
Input: [[1,3], [0,2], [1,3], [0,2]]
Output: true
Explanation:
The graph looks like this:
0----1
|    |
|    |
3----2
We can divide the vertices into two groups: {0, 2} and {1, 3}.

Example 2:
Input: [[1,2,3], [0,2], [0,1,3], [0,2]]
Output: false
Explanation:
The graph looks like this:
0----1
| \  |
|  \ |
3----2
We cannot find a way to divide the set of nodes into two independent subsets.


Note:

• graph will have length in range [1, 100].
• graph[i] will contain integers in range [0, graph.length - 1].
• graph[i] will not contain i or duplicate values.
• The graph is undirected: if any element j is in graph[i], then i will be in graph[j].

Related Topics:

## Solution 1. DFS

// OJ: https://leetcode.com/problems/is-graph-bipartite/
// Time: O(V + E)
// Space: O(V + E)
class Solution {
bool dfs(vector<vector<int>> &G, vector<int> &id, int u, int prev = 1) {
if (id[u]) return id[u] != prev;
id[u] = -prev;
for (int v : G[u]) {
if (!dfs(G, id, v, id[u])) return false;
}
return true;
}
public:
bool isBipartite(vector<vector<int>>& G) {
vector<int> id(G.size());
for (int i = 0; i < G.size(); ++i) {
if (id[i]) continue;
if (!dfs(G, id, i)) return false;
}
return true;
}
};


## Solution 2. BFS

// OJ: https://leetcode.com/problems/is-graph-bipartite/
// Time: O(V + E)
// Space: O(V + E)
class Solution {
public:
bool isBipartite(vector<vector<int>>& G) {
vector<int> id(G.size());
queue<int> q;
for (int i = 0; i < G.size(); ++i) {
if (id[i]) continue;
q.push(i);
id[i] = 1;
while (q.size()) {
int u = q.front();
q.pop();
for (int v : G[u]) {
if (id[v]) {
if (id[v] != -id[u]) return false;
continue;
}
id[v] = -id[u];
q.push(v);
}
}
}
return true;
}
};

• class Solution {
public boolean isBipartite(int[][] graph) {
int nodes = graph.length;
int[] colors = new int[nodes];
Arrays.fill(colors, -1);
for (int i = 0; i < nodes; i++) {
if (colors[i] == -1) {
colors[i] = 0;
Stack<Integer> stack = new Stack<Integer>();
stack.push(i);
while (!stack.isEmpty()) {
int start = stack.pop();
int color = colors[start];
int[] ends = graph[start];
for (int end : ends) {
if (colors[end] == -1) {
colors[end] = 1 - color;
stack.push(end);
} else if (colors[start] == colors[end])
return false;
}
}
}
}
return true;
}
}

############

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;
}
}

• // OJ: https://leetcode.com/problems/is-graph-bipartite/
// Time: O(V + E)
// Space: O(V + E)
class Solution {
public:
bool isBipartite(vector<vector<int>>& G) {
int N = G.size();
vector<int> id(N); // 0 unseen, 1 and -1 are different colors
function<bool(int, int)> dfs = [&](int u, int color) {
if (id[u]) return id[u] == color;
id[u] = color;
for (int v : G[u]) {
if (!dfs(v, -color)) return false;
}
return true;
};
for (int i = 0; i < N; ++i) {
if (!id[i] && !dfs(i, 1)) 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

############

class Solution(object):
def isBipartite(self, graph):
"""
:type graph: List[List[int]]
:rtype: bool
"""
visited = [0] * len(graph)# 0-not visited; 1-blue; 2-red;
for i in range(len(graph)):
if graph[i] and visited[i] == 0:
visited[i] = 1
q = collections.deque()
q.append(i)
while q:
v = q.popleft()#every point
for e in graph[v]:#every edge
if visited[e] != 0:
if visited[e] == visited[v]:
return False
else:
visited[e] = 3 - visited[v]
q.append(e)
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;
}