<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 {

    bool bfs(int start, vector<vector<int>>& g, vector<int>& currentColorStatus) {
        if (currentColorStatus[start] == 0) {
            currentColorStatus[start] = 1;
        }
        queue<int> q;
        q.push(start);

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

            for(auto& nb: g[person]) {
                if (currentColorStatus[nb] == 0) {
                    currentColorStatus[nb] = -currentColorStatus[person];
                    q.push(nb);
                } else if (currentColorStatus[nb] == currentColorStatus[person]) {
                    return false;
                }
            }
        }
        return true;
    }
public:
    bool possibleBipartition(int n, vector<vector<int>>& dislikes) {
        vector<vector<int>> graph(n+1);
        // convert dislikes to adjacent list which represent edge between two nodes
        for (auto& pair: dislikes) {
            int a = pair[0];
            int b = pair[1];
            graph[a].push_back(b);
            graph[b].push_back(a);
        }
        // 0: uncolor, 1: one type of color, -1 : oposite color of 1
        vector<int> color(n+1, 0); // 
        for (int person = 1; person<=n; person++) {
            if (!bfs(person, graph, color)) {
                return false;
            }
        }
        return true;
    }
};
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