<p>Given a list of strings <code>words</code> and a string <code>pattern</code>, return <em>a list of</em> <code>words[i]</code> <em>that match</em> <code>pattern</code>. You may return the answer in <strong>any order</strong>.</p>

<p>A word matches the pattern if there exists a permutation of letters <code>p</code> so that after replacing every letter <code>x</code> in the pattern with <code>p(x)</code>, we get the desired word.</p>

<p>Recall that a permutation of letters is a bijection from letters to letters: every letter maps to another letter, and no two letters map to the same letter.</p>

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

<pre>
<strong>Input:</strong> words = [&quot;abc&quot;,&quot;deq&quot;,&quot;mee&quot;,&quot;aqq&quot;,&quot;dkd&quot;,&quot;ccc&quot;], pattern = &quot;abb&quot;
<strong>Output:</strong> [&quot;mee&quot;,&quot;aqq&quot;]
<strong>Explanation:</strong> &quot;mee&quot; matches the pattern because there is a permutation {a -&gt; m, b -&gt; e, ...}. 
&quot;ccc&quot; does not match the pattern because {a -&gt; c, b -&gt; c, ...} is not a permutation, since a and b map to the same letter.
</pre>

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

<pre>
<strong>Input:</strong> words = [&quot;a&quot;,&quot;b&quot;,&quot;c&quot;], pattern = &quot;a&quot;
<strong>Output:</strong> [&quot;a&quot;,&quot;b&quot;,&quot;c&quot;]
</pre>

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

<ul>
	<li><code>1 &lt;= pattern.length &lt;= 20</code></li>
	<li><code>1 &lt;= words.length &lt;= 50</code></li>
	<li><code>words[i].length == pattern.length</code></li>
	<li><code>pattern</code> and <code>words[i]</code> are lowercase English letters.</li>
</ul>

class Solution:
    def findAndReplacePattern(self, words: List[str], pattern: str) -> List[str]:
        def create_map(string):
            char_map = OrderedDict()
            val = 1
            for char in string:
                if char not in char_map:
                    char_map[char] = val
                    val += 1

            result = []
            for char in string:
                result.append(char_map[char])
            
            return result

        pattern_map = create_map(pattern)

        result = []
        for word in words:
            if pattern_map == create_map(word):
                result.append(word)
        
        return result
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