<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>

from collections import defaultdict
from bisect import bisect_left

class Solution:
    def numMatchingSubseq(self, s: str, words: List[str]) -> int:
        # Preprocess s to create a dictionary of character indices
        char_indices = defaultdict(list)
        for i, char in enumerate(s):
            char_indices[char].append(i)
        
        # Function to check if a word is a subsequence of s
        def is_subsequence(word):
            prev_index = -1
            for char in word:
                if char not in char_indices:
                    return False
                indices = char_indices[char]
                # Find the first index greater than prev_index using binary search
                next_index = bisect_left(indices, prev_index + 1)
                if next_index == len(indices):
                    return False
                prev_index = indices[next_index]
            return True
        
        # Check each word in words
        count = 0
        for word in words:
            if is_subsequence(word):
                count += 1
        
        return count

# Example usage:
solution = Solution()
s = "abcde"
words = ["a", "bb", "acd", "ace"]
print(solution.numMatchingSubseq(s, words))  # Output: 3
20400