<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 isSubsequence(string& a, string& b) {
        int i=0;
        int n=b.length();
        int m=a.length();
        for(int j=0; j<n; j++) {
            while(i<m && a[i]!=b[j]) {
                i++;
            }
            if(i<m && a[i]==b[j]) {
                i++;
            }
            else {
                return false;
            }
        }

        return true;
    }
    int numMatchingSubseq(string s, vector<string>& words) {
        
        int count=0;
        unordered_set<string> match;
        unordered_set<string> tried;
        for(auto i: words) {
            if(match.count(i)) {
                count++;
            }

            else if(tried.count(i)) {
                continue;
            }

            else if(isSubsequence(s, i)) { 
                tried.insert(i);
                match.insert(i);
                count++;
            }

            else {
                tried.insert(i);
            }
        }

        return count;
    }
};
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