<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 {
public:
    int numMatchingSubseq(string s, vector<string>& words) {
        int sz = words.size();
        unordered_map<char, vector<int>> letter_to_indices;
        vector<int> pos(sz);
        for (int i=0; i<sz;i++){
            letter_to_indices[words[i][0]].push_back(i);
        }

        int matched_words = 0;
        for(int i=0; i<s.size(); i++){
            char c1 = s[i];

            vector<int> words_indices = letter_to_indices[c1];
            letter_to_indices[c1].clear();

            for(int j= 0; j< words_indices.size(); ++j){
                int jj = words_indices[j];
                string word = words[jj];

                pos[jj]++;
                if(pos[jj] == word.size()){
                    matched_words++;
                }
                else{
                    char c2 = word[pos[jj]];
                    letter_to_indices[c2].push_back(jj);
                }
            }
           
        }

        return matched_words;
    }

    
};
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