<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) {
        // a deque for each letter (total 26 letters)
        // for each deque: pair<int, int> -> <word idx, letter idx in the word>
        vector<deque<pair<int, int>>> checking_vec(26);
        int n = words.size();
        // put the word in each letter's waiting deque
        for (int i = 0; i < n; i++) {
            checking_vec[words[i][0] - 'a'].push_back(make_pair(i, 0));
        }

        int ans = 0;
        for (char c : s) {
            // now the words in this deque all match one more letter,
            // move them to their next waiting letter
            int deque_size = checking_vec[c - 'a'].size();

            for (int i = 0; i < deque_size; i++) {
                int w_idx = checking_vec[c - 'a'].front().first;
                int char_idx = checking_vec[c - 'a'].front().second;
                checking_vec[c - 'a'].pop_front();

                if (char_idx == words[w_idx].length() - 1) {
                    ans++;
                } else {
                    checking_vec[words[w_idx][char_idx + 1] - 'a'].push_back(
                        make_pair(w_idx, char_idx + 1));
                }
            }
        }
        return ans;
    }
};
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