<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) {
        ios_base::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);
        int ret=0,n=words.size();
        int next[s.length()][128];
        for (int i=97;i<128;i++) next[s.length()-1][i]=50001;
        for (int i=s.length()-2;i>=0;i--){
            for (int j=97;j<128;j++){
                next[i][j]=next[i+1][j];
                if (s[i+1]==j) next[i][j]=i+1;
            }
        }
        for (int k=0;k<n;k++){
            int i=0,j=0,n1=s.length(),n2=words[k].length();
            if (s[0]==words[k][0]) i=0;
            else i=next[0][words[k][0]];
            j=1;
            while (i<n1 && j<n2){
                i=next[i][words[k][j]];
                j++;
            }
            if (j==n2 && i<n1) ret++;
        }
        return ret;
    }
};
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