<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>& b) {
        int ans = 0;
        int n = s.size(), n1 = b.size();
        if(n == 1) {

        }
        vector<vector<int>> a(n, vector<int> (26, -1));
        // a[n - 1][s[n - 1] - 'a'] = ;
        vector<int> pos(26, -1);
        a[n - 2][s[n - 1] - 'a'] = n - 1;
        for(int i = n - 2; i >= 0; i--) {
            if(i - 1 >= 0) {
                a[i - 1][s[i] - 'a'] = i;
                for(int j = 0; j < 26; j++) {
                    if((char)(j + 'a') != s[i]) {
                        a[i - 1][j] = a[i][j];
                    }
                }
            }
            pos[s[i] - 'a'] = i;
        }
        // for(int i = 0; i < 26; i++) {
        //     cout << a[0][i] << " ";
        // }
        // cout << "\n";
        for(int i = 0; i < n1; i++) {
            int n3 = b[i].size();
            int j = 1;
            int st = pos[b[i][0] - 'a'];
            if(st != -1) {
                while(j < n3 && st != -1) {
                    // if(j + 1 == n3) {
                        
                    // }
                    // else {
                        st = a[st][b[i][j++] - 'a'];
                    // }
                }
            }
            if(j == n3 && st != -1) {ans++;}
        }
        return ans;
    }
};
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