couples-holding-hands

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python3

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Algorithms

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

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>ith</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>.

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

&nbsp;
Example 1:

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

Example 2:

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

&nbsp;
Constraints:

<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 unique.</li>
</ul>

minSwapsCouples

Couples Holding Hands

First Missing Positive

first-missing-positive

缺失的第一个正数

Missing Number

missing-number

丢失的数字

K-Similar Strings

k-similar-strings

相似度为 K 的字符串

Greedy

Depth-First Search

Breadth-First Search

Union Find

Graph

<code>n</code> 对情侣坐在连续排列的 <code>2n</code>&nbsp;个座位上，想要牵到对方的手。

人和座位由一个整数数组 <code>row</code> 表示，其中 <code>row[i]</code> 是坐在第 <code>i </code>个座位上的人的 ID。情侣们按顺序编号，第一对是&nbsp;<code>(0, 1)</code>，第二对是&nbsp;<code>(2, 3)</code>，以此类推，最后一对是&nbsp;<code>(2n-2, 2n-1)</code>。

返回 最少交换座位的次数，以便每对情侣可以并肩坐在一起。 每次交换可选择任意两人，让他们站起来交换座位。

&nbsp;

示例 1:

<pre>
输入: row = [0,2,1,3]
输出: 1
解释: 只需要交换row[1]和row[2]的位置即可。
</pre>

示例 2:

<pre>
输入: row = [3,2,0,1]
输出: 0
解释: 无需交换座位，所有的情侣都已经可以手牵手了。
</pre>

&nbsp;

提示:

<ul>
	<li><code>2n == row.length</code></li>
	<li><code>2 &lt;= n &lt;= 30</code></li>
	<li><code>n</code>&nbsp;是偶数</li>
	<li><code>0 &lt;= row[i] &lt; 2n</code></li>
	<li><code>row</code>&nbsp;中所有元素均无重复</li>
</ul>