<p>We want to split a group of <code>n</code> people (labeled from <code>1</code> to <code>n</code>) into two groups of <strong>any size</strong>. Each person may dislike some other people, and they should not go into the same group.</p>

<p>Given the integer <code>n</code> and the array <code>dislikes</code> where <code>dislikes[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that the person labeled <code>a<sub>i</sub></code> does not like the person labeled <code>b<sub>i</sub></code>, return <code>true</code> <em>if it is possible to split everyone into two groups in this way</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 4, dislikes = [[1,2],[1,3],[2,4]]
<strong>Output:</strong> true
<strong>Explanation:</strong> The first group has [1,4], and the second group has [2,3].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 3, dislikes = [[1,2],[1,3],[2,3]]
<strong>Output:</strong> false
<strong>Explanation:</strong> We need at least 3 groups to divide them. We cannot put them in two groups.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 2000</code></li>
	<li><code>0 &lt;= dislikes.length &lt;= 10<sup>4</sup></code></li>
	<li><code>dislikes[i].length == 2</code></li>
	<li><code>1 &lt;= a<sub>i</sub> &lt; b<sub>i</sub> &lt;= n</code></li>
	<li>All the pairs of <code>dislikes</code> are <strong>unique</strong>.</li>
</ul>

class Solution {
public:

    bool BFS(vector<vector<int>>& list,vector<int>&g,int source){
        queue<int>q;
        q.push(source);
        g[source]=1;

        while(!q.empty()){
            int node=q.front();
            q.pop();

            for(int i=0;i<list[node].size();i++){
                int newNode=list[node][i];
                if(g[newNode]!=0 && g[newNode]==g[node]){
                    return 0;
                }else if(g[newNode]==0){
                    g[newNode]=g[node]*-1;
                    q.push(newNode);
                }
            }
        }

        return 1;
    }

    bool possibleBipartition(int n, vector<vector<int>>& d) {
        vector<int>g(n,0);
        vector<vector<int>> list(n);

        for(int i=0;i<d.size();i++){
            list[d[i][0]-1].push_back(d[i][1]-1);
            list[d[i][1]-1].push_back(d[i][0]-1);
        }

        bool ans=1;
        for(int i=0;i<n;i++){
            if(g[i]==0){
                ans= ans & BFS(list,g,i);
            }
        }

        return ans;
    }
};
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