<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]:
        result = []


        def match(hasher, firstLet, secondLet):
            if hasher[firstLet] != secondLet:
                return False
            return True
        
        def checker(wdLet, paLet):
            matching = {}
            pmatching = {}
            flag = 1
            for i in range(len(wd)):
                wdLet = wd[i]
                paLet = pattern[i]
                if wdLet in matching and not match(matching, wdLet, paLet):
                        flag = 0
                        break
                elif paLet in pmatching and not match(pmatching,paLet, wdLet):
                        flag = 0
                        break
                elif wdLet not in matching and paLet not in pmatching:
                    matching[wdLet] = paLet
                    pmatching[paLet] = wdLet
            if flag:
                return True
            return False

        for wd in words:
            if checker(wd, pattern):
                result.append(wd)
           

        return result
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