<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:
    bool isSubSeq(string &word, vector<vector<int>> &dp_pos) {
        int c_pos = 0;
        int s_len = dp_pos[0].size();
        for (int i = 0; i < word.size(); i++) {
            if (c_pos >= s_len || dp_pos[word[i] - 'a'][c_pos] == -1) return false;
            c_pos = dp_pos[word[i] - 'a'][c_pos] + 1;
        }
        return true;
    }
    int numMatchingSubseq(string s, vector<string>& words) {
        int ans = 0;
        int n = words.size();
        int s_len = s.length();
        vector<vector<int>> dp_pos(26, vector<int> (s_len, -1));
        for (char ch = 'a'; ch <= 'z'; ch++) {
            dp_pos[ch - 'a'][s_len - 1] = (s[s_len - 1] == ch) ? (s_len - 1) : -1;
            for (int i = s_len - 2; i >= 0; i--) {
                if (s[i] == ch) {
                    dp_pos[ch - 'a'][i] = i;
                } else {
                    dp_pos[ch - 'a'][i] = dp_pos[ch - 'a'][i + 1];
                }
            }
        }
        for (int i = 0; i < n; i++) {
            if (isSubSeq(words[i], dp_pos)) ans++;
        }
        return ans;
    }
};
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