Check whether a given graph is Bipartite or not (original) (raw)
Last Updated : 23 Jul, 2025
Given a graph with **V vertices numbered from **0 to V-1 and a list of **edges, determine whether the graph is **bipartite or not.
**Note: A **bipartite graph is a type of graph where the set of vertices can be divided into **two disjoint sets, say **U and **V, such that **every edge connects a vertex in U to a vertex in V, there are **no edges between vertices within the same set.
**Example:
**Input: V = 4, edges[][]= [[0, 1], [0, 2], [1, 2], [2, 3]]
**Output: false
**Explanation:
node 1 and node 2 have same color while coloring
The graph is not bipartite because no matter how we try to color the nodes using two colors, there exists a cycle of odd length (like 1–2–0–1), which leads to a situation where two adjacent nodes end up with the same color. This violates the bipartite condition, which requires that no two connected nodes share the same color.
**Input: _V = 4, edges[][] = [[0, 1], [1, 2], [2, 3]]
**Output: true
**Explanation:
The given graph can be colored in two colors so, it is a bipartite graph.
Table of Content
**Using Breadth-First Search (BFS)
Checking if a graph is bipartite is like trying to color the graph using only two colors, so that no two adjacent vertices have the same color. One approach is to check whether the graph is 2-colorable or not using backtracking algorithm m coloring problem.
A common and efficient way to solve this is by using **Breadth-First Search (BFS). The idea is to traverse the graph level by level and assign colors alternately to the vertices as we proceed.
Step-by-step approach:
- Start BFS from any uncolored vertex and assign it color **0.
- For each vertex, color its uncolored neighbors with the opposite color (**1 if current is **0, and vice versa)
- Check if a neighbor already has the same color as the current vertex, return **false (graph is not bipartite).
- If BFS completes without any conflicts, return **true (graph is bipartite).
Below is the implementation of the above approach:
C++ `
//Driver Code Starts #include <bits/stdc++.h> using namespace std;
vector<vector> constructadj(int V, vector<vector> &edges){
vector<vector<int>> adj(V);
for(auto it:edges){
adj[it[0]].push_back(it[1]);
adj[it[1]].push_back(it[0]);
}
return adj;} // Function to check if the graph is Bipartite using BFS //Driver Code Ends
bool isBipartite(int V, vector<vector> &edges) {
// Vector to store colors of vertices.
// Initialize all as -1 (uncolored)
vector<int> color(V, -1);
//create adjacency list
vector<vector<int>> adj = constructadj(V,edges);
// Queue for BFS
queue<int> q;
// Iterate through all vertices to handle disconnected graphs
for(int i = 0; i < V; i++) {
// If the vertex is uncolored, start BFS from it
if(color[i] == -1) {
// Assign first color (0) to the starting vertex
color[i] = 0;
q.push(i);
// Perform BFS
while(!q.empty()) {
int u = q.front();
q.pop();
// Traverse all adjacent vertices
for(auto &v : adj[u]) {
// If the adjacent vertex is uncolored,
// assign alternate color
if(color[v] == -1) {
color[v] = 1 - color[u];
q.push(v);
}
// If the adjacent vertex has the same color,
// graph is not bipartite
else if(color[v] == color[u]) {
return false;
}
}
}
}
}
// If no conflicts in coloring, graph is bipartite//Driver Code Starts return true; }
int main() { int V = 4; vector<vector> edges = {{0, 1}, {0, 2}, {1, 2}, {2, 3}}; if(isBipartite(V, edges)) cout << "true"; else cout << "false";
return 0;}
//Driver Code Ends
Java
//Driver Code Starts import java.util.*;
class GfG {
// Function to construct the adjacency list from edges
static ArrayList<ArrayList<Integer>> constructadj(int V, int[][] edges) {
ArrayList<ArrayList<Integer>> adj = new ArrayList<>();
for (int i = 0; i < V; i++) {
adj.add(new ArrayList<>());
}
for (int[] edge : edges) {
int u = edge[0];
int v = edge[1];
adj.get(u).add(v);
adj.get(v).add(u);
}
return adj;
}
// Function to check if the graph is Bipartite using BFS//Driver Code Ends
static boolean isBipartite(int V,int[][] edges) {
int[] color = new int[V];
Arrays.fill(color, -1);
// create adjacency list
ArrayList<ArrayList<Integer>> adj = constructadj(V,edges);
for (int i = 0; i < V; i++) {
if (color[i] == -1) {
Queue<Integer> q = new LinkedList<>();
color[i] = 0;
q.offer(i);
while (!q.isEmpty()) {
int u = q.poll();
for (int v : adj.get(u)) {
if (color[v] == -1) {
color[v] = 1 - color[u];
q.offer(v);
} else if (color[v] == color[u]) {
return false; // Conflict found
}
}
}
}//Driver Code Starts }
return true;
}
public static void main(String[] args) {
int V = 4;
// Edges of the graph
int[][] edges = {{0, 1}, {0, 2}, {1, 2}, {2, 3}};
// Check if the graph is bipartite
System.out.println(isBipartite(V, edges));
}}
//Driver Code Ends
Python
#Driver Code Starts from collections import deque
Function to construct the adjacency list from edges
def constructadj(V, edges):
adj = [[] for _ in range(V)]
for edge in edges:
u, v = edge
adj[u].append(v)
adj[v].append(u)
return adj
Function to check if the graph is Bipartite using BFS
#Driver Code Ends
def isBipartite(V, adj): # Initialize all as uncolored color = [-1] * V
# create adjacency list
adj = constructadj(V,edges)
for i in range(V):
if color[i] == -1:
color[i] = 0
q = deque([i])
while q:
u = q.popleft()
for v in adj[u]:
if color[v] == -1:
color[v] = 1 - color[u]
q.append(v)
elif color[v] == color[u]:
return False # Conflict found#Driver Code Starts
# No conflict, graph is bipartite
return True if name == "main": V = 4 edges = [[0, 1], [0, 2], [1, 2], [2, 3]]
print("true" if isBipartite(V, edges) else "false")#Driver Code Ends
C#
//Driver Code Starts using System; using System.Collections.Generic;
class GfG {
public static List<List<int>> constructadj(int V, List<List<int>> edges)
{
List<List<int>> adj = new List<List<int>>();
for (int i = 0; i < V; i++)
adj.Add(new List<int>());
foreach (var edge in edges)
{
int u = edge[0];
int v = edge[1];
adj[u].Add(v);
adj[v].Add(u);
}
return adj;
}//Driver Code Ends
// Function to check if the graph is Bipartite using BFS
public static bool IsBipartite(int V,
List<List<int> > edges){
int[] color = new int[V];
// create adjacency list
List<List<int>> adj = constructadj(V, edges);
// Initialize all as -1 (uncolored)
Array.Fill(color, -1);
// Iterate through all vertices to handle
// disconnected graphs
for (int i = 0; i < V; i++) {
// If the vertex is uncolored, start BFS from it
if (color[i] == -1) {
// Assign first color (0)
color[i] = 0;
Queue<int> q = new Queue<int>();
q.Enqueue(i);
// Perform BFS
while (q.Count > 0) {
int u = q.Dequeue();
// Traverse all adjacent vertices
foreach(int v in adj[u]){
// If the adjacent vertex is
// uncolored, assign alternate color
if (color[v] == -1) {
color[v] = 1 - color[u];
q.Enqueue(v);
}
// If the adjacent vertex has the
// same color, graph is not
// bipartite
else if (color[v] == color[u]) {
return false;
}
}
}
}
}
// If no conflicts in coloring, graph is bipartite//Driver Code Starts return true; }
static void Main(){
int V = 4;
List<List<int>> edges = new List<List<int>> {
new List<int>{0, 1},
new List<int>{0, 2},
new List<int>{1, 2},
new List<int>{2, 3}
};
if (IsBipartite(V, edges))
Console.WriteLine("true");
else
Console.WriteLine("false");
}} //Driver Code Ends
JavaScript
//Driver Code Starts // Function to construct adjacency list from edges function constructadj(V, edges) { const adj = Array.from({ length: V }, () => []);
for (const [u, v] of edges) {
adj[u].push(v);
adj[v].push(u); // undirected graph
}
return adj;}
//Driver Code Ends
function isBipartite(V, edges){
// Initialize all as -1 (uncolored)
const color = Array(V).fill(-1);
// create adjacency list
let adj = constructadj(V,edges);
// Iterate through all vertices to handle disconnected
// graphs
for (let i = 0; i < V; i++) {
// If the vertex is uncolored, start BFS from it
if (color[i] === -1) {
// Assign first color (0)
color[i] = 0;
const queue = [ i ];
// Perform BFS
while (queue.length > 0) {
const u = queue.shift();
// Traverse all adjacent vertices
for (let v of adj[u]) {
// If the adjacent vertex is uncolored,
// assign alternate color
if (color[v] === -1) {
color[v] = 1 - color[u];
queue.push(v); // Push to queue
}
// If the adjacent vertex has the same
// color, graph is not bipartite
else if (color[v] === color[u]) {
return false;
}
}
}
}
}
// If no conflicts in coloring, graph is bipartite//Driver Code Starts return true; }
// Driver Code const V = 4; const adj = Array.from({length : V}, () => []);
let edges = [[0, 1], [0, 2], [1, 2], [2, 3]];
console.log(isBipartite(V, edges)); //Driver Code Ends
`
**Time Complexity: O(V + E), where V is the number of vertices and E is the number of edges. This is because BFS explores each vertex and edge exactly once.
**Auxiliary Space: O(V), The queue used in BFS, which can hold up to V vertices and The color array (or map), which stores the color for each vertex, We do not count the adjacency list in auxiliary space as it is necessary for representing the input graph.
**Using Depth-First Search (DFS)
We can also check if a graph is bipartite using **Depth-First Search (DFS). We need to color the graph with two colors such that no two adjacent vertices share the same color. We start from any uncolored vertex, assigning it a color (e.g., color 0). As we explore each vertex, we recursively color its uncolored neighbors with the another color. If we ever find a neighbor that shares the same color as the current vertex, we can simply conclude that the graph is not bipartite. If there is no conflict found after the traversal then the given graph is bipartite.
For implementation of DFS approach please refer to this article "Check if a given graph is Bipartite using DFS".
**Related Article: