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
func possibleBipartition(n int, dislikes [][]int) bool {
	g := make([][]int, n)
	for _, e := range dislikes {
		a, b := e[0]-1, e[1]-1
		g[a] = append(g[a], b)
		g[b] = append(g[b], a)
	}
	color := make([]int, n)
	var dfs func(int, int) bool
	dfs = func(i, c int) bool {
		color[i] = c
		for _, j := range g[i] {
			if color[j] == c {
				return false
			}
			if color[j] == 0 && !dfs(j, 3-c) {
				return false
			}
		}
		return true
	}
	for i, c := range color {
		if c == 0 && !dfs(i, 1) {
			return false
		}
	}
	return true
}