<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 {
public:
    int compa(string s, string pat) {
        vector<int> a(27, -1);
        vector<int> b(27,-1);
        int i = 0;
        //int ss =0,spat=0;
        while (i < s.size()) {
            //check in a
            if (a[pat[i] - 'a'] != s[i] && a[pat[i] - 'a'] != -1) {
                return 0;
            }

            //check in b
            if (b[s[i] - 'a'] != pat[i] && b[s[i] - 'a'] != -1) {
                return 0;
            }
            a[pat[i] - 'a'] = s[i];
            b[s[i] - 'a'] = pat[i];

            i++;
        }
        return 1;
    }

    vector<string> findAndReplacePattern(vector<string>& wor, string pat) {
        vector<string> vs;
        for (int i = 0; i < wor.size(); i++) {
            int cmp = compa(wor[i], pat);
            if (cmp)
                vs.emplace_back(wor[i]);
        }
        return vs;
    }
};
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