<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>

#pragma GCC optimize("O3")

class Solution {
public:
    int numMatchingSubseq(string s, vector<string>& words) {
        vector<std::array<unsigned short, 26>> closest_right{s.size()};
        std::fill(closest_right[s.size() - 1].begin(), closest_right[s.size() - 1].end(), USHRT_MAX);
        for (int i = s.size() - 1; i > 0; --i) {
            closest_right[i - 1] = closest_right[i];
            closest_right[i - 1][s[i] - 'a'] = i;
        }
        int ans = 0;
        for (auto& word : words) {
            int start = 0;
            if (s[0] != word[0]) {
                start = closest_right[0][word[0] - 'a'];
            }
            if (start == USHRT_MAX) {
                continue;
            }
            ans += 1;
            for (int i = 1; i < word.size(); ++i) {
                start = closest_right[start][word[i] - 'a'];
                if (start == USHRT_MAX) {
                    ans -= 1;
                    break;
                }
            }
        }
        return ans;
    }
};
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