5.Possible Bipartition

Bipartite Graph Check

A Bipartite Graph is a graph whose vertices can be divided into two independent sets, U and V such that every edge (u, v) either connects a vertex from U to V or a vertex from V to U.

Question:

Given a set of N people (numbered 1, 2, ..., N), we would like to split everyone into two groups of any size. Each person may dislike some other people, and they should not go into the same group.

Formally, if dislikes[i] = [a, b], it means it is not allowed to put the people numbered a and b into the same group.

Return true if and only if it is possible to split everyone into two groups in this way.

Example 1:
Input: N = 4, dislikes = [[1,2],[1,3],[2,4]]
Output: true
Explanation: group1 [1,4], group2 [2,3]

Example 2:
Input: N = 3, dislikes = [[1,2],[1,3],[2,3]]
Output: false

Example 3:
Input: N = 5, dislikes = [[1,2],[2,3],[3,4],[4,5],[1,5]]
Output: false

Solution : (Check for Bipartite Graph using Color)

class Solution
{
public:
    bool checkBipartite(int itr, vector<int> v[], vector<bool> &vs, vector<int> &col)
    {

        queue<int> q;
        q.push(itr);
        vs[itr] = 1;
        col[itr] = 1;

        while (!q.empty())
        {
            int node = q.front();
            int colr = col[node];
            q.pop();
            
            for (int i = 0; i < v[node].size(); i++)
            {
                if (vs[v[node][i]] == 0)
                {
                    q.push(v[node][i]);
                    vs[v[node][i]] = 1;
                    col[v[node][i]] = 1 - colr;
                }
                else
                {
                    if (col[v[node][i]] == colr)
                    {
                        return false;
                    }
                }
            }
        }
        return true;
    }

    bool possibleBipartition(int n, vector<vector<int>> &dislikes)
    {

        vector<int> v[n + 1];
        vector<bool> vs(n + 1);
        vector<int> col(n + 1, -1);
        for (int i = 0; i < dislikes.size(); i++)
        {
            v[dislikes[i][0]].push_back(dislikes[i][1]);
            v[dislikes[i][1]].push_back(dislikes[i][0]);
        }

        for (int i = 1; i <= n; i++)
        {
            if (vs[i] == 0)
            {
                if (!checkBipartite(i, v, vs, col))
                {
                    return false;
                }
            }
        }

        return true;
    }
};