<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>

vector<vector<int>> dp(50001, vector<int>(51, 0));
class Solution {
public:
    
    
    int numMatchingSubseq(string s, vector<string>& words){
        
        vector<vector<int>> charIndexes(26);
        for(int i = 0;i < s.size(); i++){
            charIndexes[s[i]-'a'].push_back(i);
        }
        
        int count = 0;
        
        for(int i = 0;i< words.size();i++){
            bool isSubseq = true;
            int lastCharIndex = -1;
            for(char w : words[i]){
                auto it = upper_bound(charIndexes[w - 'a'].begin(), charIndexes[w - 'a'].end(), lastCharIndex);
                if(it==charIndexes[w - 'a'].end()){
                    isSubseq = false;
                    break;
                }
                else{
                    lastCharIndex = *it;
                }
                
            }
            
            if(isSubseq){
                count++;
            }
            
            
        }
        
        
        return count;
        
    }
    
    
};    
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