<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:
        from collections import defaultdict
        from bisect import bisect_right

        # Step 1: Create a map of character to all its positions (indices) in the string s
        char_indices = defaultdict(list)
        for idx, char in enumerate(s):
            char_indices[char].append(idx)

        def is_subsequence(word):
            # Previous index in 's' where a matching character was found
            prev_idx = -1
            for char in word:
                # If the character does not exist in `s`, word cannot be a subsequence
                if char not in char_indices:
                    return False

                # Step 2: Find the next index in `s` after the character at `prev_idx`
                idx_list = char_indices[char]
                pos = bisect_right(idx_list, prev_idx)
                if pos >= len(idx_list):
                    return False

                # Update prev_idx to the found index
                prev_idx = idx_list[pos]

            return True

        # Step 3: Count the words that are subsequences of `s`
        count = 0
        for word in words:
            if is_subsequence(word):
                count += 1

        return count
20000