<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 UnionFind{
    public:
    vector<int> parent, size;
    int n;
    UnionFind(int a)
    {
        n = a;
        for(int i = 0; i < n; i++)
        {
            parent.push_back(i);
            size.push_back(1);
        }
    }
    int find(int &a)
    {
        if(parent[a] != a)
        {
            parent[a] = find(parent[a]);
        }
        return parent[a];
    }
    bool connect(int &a, int &b)
    {
        int roota = find(a);
        int rootb = find(b);
        if(roota == rootb)
            return false;
        if(size[roota] > size[rootb])
        {
            parent[rootb] = roota;
            size[roota] += size[rootb];
        }
        else{
            parent[roota] = rootb;
            size[rootb] += size[roota];
        }
        return true;
    }
};
class Solution {
public:
    bool possibleBipartitionBipartite(int n, vector<vector<int>>& dislikes)
    {
        vector<vector<int>> G(n + 1);
        for(auto edge : dislikes)
        {
            G[edge[0]].push_back(edge[1]);
            G[edge[1]].push_back(edge[0]);
        }
        vector<int> color(n + 1, -1);
        for(int i = 1; i <= n; i++)
        {
            if(color[i] != -1)
                continue;
            queue<int> q;
            q.push(i);
            color[i] = 0;
            while(q.empty() == false)
            {
                int f = q.front();
                q.pop();
                for(auto neighbor : G[f])
                {
                    if(color[neighbor] == -1)
                    {
                        color[neighbor] = 1 - color[f];
                        q.push(neighbor);
                    }
                    else{
                        if(color[neighbor] == color[f])
                            return false;
                    }
                }
            }
        }
        return true;
    } 
    bool possibleBipartition(int n, vector<vector<int>>& dislikes) {
        return possibleBipartitionUnionFind(n, dislikes);
    }
    bool possibleBipartitionUnionFind(int n, vector<vector<int>>& dislikes) {
        vector<vector<int>> G(n + 1);
        for(auto edge : dislikes)
        {
            G[edge[0]].push_back(edge[1]);
            G[edge[1]].push_back(edge[0]);
        }
        UnionFind uf(n + 1);
        for(int node = 1; node <= n; node++)
        {
            for(auto neighbor : G[node])
            {
                if(uf.find(node) == uf.find(neighbor))
                    return false;
                uf.connect(G[node][0], neighbor);
            }
        }
        return true;
    }
};
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