Possible Bipartition
<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>
Depth-First Search; Breadth-First Search; Union Find; Graph
class Solution {
public:
    bool possibleBipartition(int n, vector<vector<int>>& dislikes) {
        unordered_map<int, vector<int>> g;
        for (auto& e : dislikes) {
            int a = e[0] - 1, b = e[1] - 1;
            g[a].push_back(b);
            g[b].push_back(a);
        }
        vector<int> color(n);
        function<bool(int, int)> dfs = [&](int i, int c) -> bool {
            color[i] = c;
            for (int j : g[i]) {
                if (!color[j] && !dfs(j, 3 - c)) return false;
                if (color[j] == c) return false;
            }
            return true;
        };
        for (int i = 0; i < n; ++i) {
            if (!color[i] && !dfs(i, 1)) return false;
        }
        return true;
    }
};