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
class Solution {
    private int[] p;

    public int minSwapsCouples(int[] row) {
        int n = row.length >> 1;
        p = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
        }
        for (int i = 0; i < n << 1; i += 2) {
            int a = row[i] >> 1, b = row[i + 1] >> 1;
            p[find(a)] = find(b);
        }
        int ans = n;
        for (int i = 0; i < n; ++i) {
            if (i == find(i)) {
                --ans;
            }
        }
        return ans;
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}