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>