Couples Holding Hands
<p>There are <code>n</code> couples sitting in <code>2n</code> seats arranged in a row and want to hold hands.</p>

<p>The people and seats are represented by an integer array <code>row</code> where <code>row[i]</code> is the ID of the person sitting in the <code>i<sup>th</sup></code> seat. The couples are numbered in order, the first couple being <code>(0, 1)</code>, the second couple being <code>(2, 3)</code>, and so on with the last couple being <code>(2n - 2, 2n - 1)</code>.</p>

<p>Return <em>the minimum number of swaps so that every couple is sitting side by side</em>. A swap consists of choosing any two people, then they stand up and switch seats.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> row = [0,2,1,3]
<strong>Output:</strong> 1
<strong>Explanation:</strong> We only need to swap the second (row[1]) and third (row[2]) person.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> row = [3,2,0,1]
<strong>Output:</strong> 0
<strong>Explanation:</strong> All couples are already seated side by side.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2n == row.length</code></li>
	<li><code>2 &lt;= n &lt;= 30</code></li>
	<li><code>n</code> is even.</li>
	<li><code>0 &lt;= row[i] &lt; 2n</code></li>
	<li>All the elements of <code>row</code> are <strong>unique</strong>.</li>
</ul>
Greedy; Depth-First Search; Breadth-First Search; Union Find; Graph
func minSwapsCouples(row []int) int {
	n := len(row) >> 1
	p := make([]int, n)
	for i := range p {
		p[i] = i
	}
	var find func(int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	for i := 0; i < n<<1; i += 2 {
		a, b := row[i]>>1, row[i+1]>>1
		p[find(a)] = find(b)
	}
	ans := n
	for i := range p {
		if find(i) == i {
			ans--
		}
	}
	return ans
}