bipartite_graph.cpp

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].






// Finding bipartite using graph coloring concept

// BFS
// TC -> O(V + E)
// SC -> O(V)

class Solution {
public:
    
    bool bipartiteBFS(int s, vector<vector<int>> &graph, vector <int> &color) {
        queue <int> q;
        q.push(s);
        color[s] = 1; // initial coloring
        
        while (!q.empty()) {
            int node = q.front();
            q.pop();
            
            for (auto &x: graph[node]) {
                if (color[x] == -1) {
                    q.push(x);
                    color[x] = !(color[node]);  // giving opposite color
                } else {
                    if (color[x] == color[node]) return false;   // not bipartite
                }
            }
        }
        
        return true;    // bipartite
    }
    
    bool isBipartite(vector<vector<int>>& graph) {
        int n = graph.size();
        
        vector <int> color(n, -1);
        
        for (int i = 0; i < n; i++) {
            if (color[i] == -1) {
                if (!bipartiteBFS(i, graph, color)) return false;
            }
        }
        
        return true;
    }
};


















// DFS
// TC -> O(V + E)
// SC -> O(V)

class Solution {
public:
    
    bool bipartiteDFS(int s, vector<vector<int>> &graph, vector <int> &color) {
        if (color[s] == -1) color[s] = 1; // initial coloring
        
        for (auto &x: graph[s]) {
            if (color[x] == -1) {
                color[x] = !(color[s]);  // coloring opposite
                if (!bipartiteDFS(x, graph, color)) return false;
            } else {
                if (color[x] == color[s]) return false;  // not bipartite
            }
        }

        return true; // bipartite
    }
    
    bool isBipartite(vector<vector<int>>& graph) {
        int n = graph.size();
        
        vector <int> color(n, -1);
        
        for (int i = 0; i < n; i++) {
            if (color[i] == -1) {
                if (!bipartiteDFS(i, graph, color)) return false;
            }
        }
        
        return true;
    }
};