<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 n = s.length();
//         int cnt = 0;
//         for(int k=0; k<words.size(); k++) {
//             string t = words[k];
//             int m = t.length();
//             vector<vector<int>> dp(m+1, vector<int> (n+1, 0));
            
//             for(int i=1; i<=m; i++) {
//                 for(int j=1; j<=n; j++) {
//                     if(t[i-1] == s[j-1]) {
//                         dp[i][j] = 1 + dp[i-1][j-1];
//                     }
//                     else {
//                         dp[i][j] = dp[i][j-1];
//                     }
//                 }
//             }
//             if(dp[m][n] == m)
//                 cnt++;
//         }
//         return cnt;
//     }
// };

        array<vector<string_view>, 26> cursor;
        for (auto& word : words)
            cursor[word[0] - 'a'].push_back(word);
        int ret = 0;
        for (auto c : s) {
            auto cur = move(cursor[c - 'a']);
            for (auto next : cur)
                if (1 == size(next))
                    ++ret;
                else
                    cursor[next[1] - 'a'].push_back(next.substr(1));
        }
        return ret;
    }
};
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