<p>Given a string <code>s</code> and an array of strings <code>words</code>, return <em>the number of</em> <code>words[i]</code> <em>that is a subsequence of</em> <code>s</code>.</p>

<p>A <strong>subsequence</strong> of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.</p>

<ul>
	<li>For example, <code>&quot;ace&quot;</code> is a subsequence of <code>&quot;abcde&quot;</code>.</li>
</ul>

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

<pre>
<strong>Input:</strong> s = &quot;abcde&quot;, words = [&quot;a&quot;,&quot;bb&quot;,&quot;acd&quot;,&quot;ace&quot;]
<strong>Output:</strong> 3
<strong>Explanation:</strong> There are three strings in words that are a subsequence of s: &quot;a&quot;, &quot;acd&quot;, &quot;ace&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;dsahjpjauf&quot;, words = [&quot;ahjpjau&quot;,&quot;ja&quot;,&quot;ahbwzgqnuk&quot;,&quot;tnmlanowax&quot;]
<strong>Output:</strong> 2
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= words.length &lt;= 5000</code></li>
	<li><code>1 &lt;= words[i].length &lt;= 50</code></li>
	<li><code>s</code> and <code>words[i]</code> consist of only lowercase English letters.</li>
</ul>

class Solution:
    def numMatchingSubseq(self, s: str, words: List[str]) -> int:
        count = 0
        hashMap = {}
        n = len(s)
        for i in range(n):
            if s[i] in hashMap:
                hashMap[s[i]].append(i)
            else:
                hashMap[s[i]] = [i]
        # print(hashMap)
        wordsFreq = {}
        for word in words:
            if word in wordsFreq:
                wordsFreq[word] += 1
            else:
                wordsFreq[word] = 1

        for word in wordsFreq:
            i = 0
            j = 0
            m = len(word)
            prev = -1
            while i<n and j<m:
                matchLetter = word[j]
                if matchLetter in hashMap:
                    tried = 0
                    for poss in hashMap[matchLetter]:
                        if poss > prev:
                            i=poss + 1
                            j+=1
                            prev = poss
                            break
                        else:
                            tried += 1
                    if tried == len(hashMap[matchLetter]):
                        break
                        
                else:
                    break
            if j == m:
                count += wordsFreq[word]
        return count
20500