<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 dfs(int node,vector<int>adj[],vector<int>& vis,vector<int> &par,vector<int> &count){
    //     for(auto it:adj[node]){
    //         if(!vis[it]){
    //             vis[it]=1;
    //             par[it]=node;
    //             count[it]=count[node]+1;
    //             if(!dfs(it,adj,vis,par,count))return false;
    //         }
    //         else if(par[it]!=node){
    //             if(abs(count[it]-count[node])&1){
    //                 return false;
    //             }
    //         }
    //     }
    //     return true;
    // }
    // bool possibleBipartition(int n, vector<vector<int>>& dislikes) {
    //     vector<int>adj[n+1];
    //     for(int i=0;i<dislikes.size();i++){
    //          adj[dislikes[i][0]].push_back(dislikes[i][1]);
    //          adj[dislikes[i][1]].push_back(dislikes[i][0]);
    //     }
    //     vector<int>vis(n+1,0),par(n+1,-1);
    //     vector<int>count(n+1,0);
    //     for(int i=1;i<=n;i++){
    //         if(!vis[i]){
    //             if(!dfs(i,adj,vis,par,count)){
    //                 return false;
    //             }
    //         }
    //     }
    //     return true;
    // }
    bool possibleBipartition(int n, vector<vector<int>>& dislikes) {
    vector<int> adj[n + 1];
    for (const auto& dislike : dislikes) {
        adj[dislike[0]].push_back(dislike[1]);
        adj[dislike[1]].push_back(dislike[0]);
    }

    vector<int> color(n + 1, -1); // -1 means uncolored, 0 and 1 are two colors

    for (int i = 1; i <= n; ++i) {
        if (color[i] == -1) {
            // Start BFS from node i
            queue<int> q;
            q.push(i);
            color[i] = 0;

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

                for (int neighbor : adj[node]) {
                    if (color[neighbor] == -1) {
                        // Color with opposite color
                        color[neighbor] = 1 - color[node];
                        q.push(neighbor);
                    } else if (color[neighbor] == color[node]) {
                        // Conflict in colors
                        return false;
                    }
                }
            }
        }
    }

    return true;
}
};
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