determine-if-two-strings-are-close

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Algorithms

Two strings are considered close if you can attain one from the other using the following operations:

<ul>
	<li>Operation 1: Swap any two existing characters.

	<ul>
		<li>For example, <code>abcde -&gt; aecdb</code></li>
	</ul>
	</li>
	<li>Operation 2: Transform every occurrence of one existing character into another existing character, and do the same with the other character.
	<ul>
		<li>For example, <code>aacabb -&gt; bbcbaa</code> (all <code>a</code>&#39;s turn into <code>b</code>&#39;s, and all <code>b</code>&#39;s turn into <code>a</code>&#39;s)</li>
	</ul>
	</li>
</ul>

You can use the operations on either string as many times as necessary.

Given two strings, <code>word1</code> and <code>word2</code>, return <code>true</code> if <code>word1</code> and <code>word2</code> are close, and <code>false</code> otherwise.

&nbsp;
Example 1:

<pre>
Input: word1 = &quot;abc&quot;, word2 = &quot;bca&quot;
Output: true
Explanation: You can attain word2 from word1 in 2 operations.
Apply Operation 1: &quot;abc&quot; -&gt; &quot;acb&quot;
Apply Operation 1: &quot;acb&quot; -&gt; &quot;bca&quot;
</pre>

Example 2:

<pre>
Input: word1 = &quot;a&quot;, word2 = &quot;aa&quot;
Output: false
Explanation: It is impossible to attain word2 from word1, or vice versa, in any number of operations.
</pre>

Example 3:

<pre>
Input: word1 = &quot;cabbba&quot;, word2 = &quot;abbccc&quot;
Output: true
Explanation: You can attain word2 from word1 in 3 operations.
Apply Operation 1: &quot;cabbba&quot; -&gt; &quot;caabbb&quot;
Apply Operation 2: &quot;caabbb&quot; -&gt; &quot;baaccc&quot;
Apply Operation 2: &quot;baaccc&quot; -&gt; &quot;abbccc&quot;
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= word1.length, word2.length &lt;= 105</code></li>
	<li><code>word1</code> and <code>word2</code> contain only lowercase English letters.</li>
</ul>

closeStrings

Determine if Two Strings Are Close

Buddy Strings

buddy-strings

亲密字符串

Minimum Swaps to Make Strings Equal

minimum-swaps-to-make-strings-equal

交换字符使得字符串相同

Minimum Number of Steps to Make Two Strings Anagram

minimum-number-of-steps-to-make-two-strings-anagram

制造字母异位词的最小步骤数

Hash Table

String

Counting

Sorting

如果可以使用以下操作从一个字符串得到另一个字符串，则认为两个字符串 接近 ：

<ul>
	<li>操作 1：交换任意两个 现有 字符。

	<ul>
		<li>例如，<code>abcde -&gt; aecdb</code></li>
	</ul>
	</li>
	<li>操作 2：将一个 现有 字符的每次出现转换为另一个 现有 字符，并对另一个字符执行相同的操作。
	<ul>
		<li>例如，<code>aacabb -&gt; bbcbaa</code>（所有 <code>a</code> 转化为 <code>b</code> ，而所有的 <code>b</code> 转换为 <code>a</code> ）</li>
	</ul>
	</li>
</ul>

你可以根据需要对任意一个字符串多次使用这两种操作。

给你两个字符串，<code>word1</code> 和 <code>word2</code> 。如果 <code>word1</code> 和 <code>word2</code> 接近 ，就返回 <code>true</code> ；否则，返回 <code>false</code> 。

&nbsp;

示例 1：

<pre>
输入：word1 = "abc", word2 = "bca"
输出：true
解释：2 次操作从 word1 获得 word2 。
执行操作 1："abc" -&gt; "acb"
执行操作 1："acb" -&gt; "bca"
</pre>

示例 2：

<pre>
输入：word1 = "a", word2 = "aa"
输出：false
解释：不管执行多少次操作，都无法从 word1 得到 word2 ，反之亦然。</pre>

示例 3：

<pre>
输入：word1 = "cabbba", word2 = "abbccc"
输出：true
解释：3 次操作从 word1 获得 word2 。
执行操作 1："cabbba" -&gt; "caabbb"
执行操作 2：<code>"</code>caabbb" -&gt; "baaccc"
执行操作 2："baaccc" -&gt; "abbccc"
</pre>

提示：

<ul>
	<li><code>1 &lt;= word1.length, word2.length &lt;= 105</code></li>
	<li><code>word1</code> 和 <code>word2</code> 仅包含小写英文字母</li>
</ul>