<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>& v1) {
        vector<pair<char,int>>v;
        for(int i=0;i<s.length();i++) {
            v.push_back({s[i],i});
        }
        sort(v.begin(),v.end());
        int cnt=0;
        for(int i=0;i<v1.size();i++) {
            int j=0;
            int st=-1;
            for(;j<v1[i].length();j++) {
                auto k = lower_bound(v.begin(), v.end(), make_pair(v1[i][j], st+1),
                                 [](const pair<char, int>& a, const pair<char, int>& b) {
                                     if (a.first == b.first) {
                                         return a.second < b.second;
                                     }
                                     return a.first < b.first;
                                 });
                int pos = k-v.begin();
                if (pos>=v.size()) {
                    break;
                }
                if(v[pos].first!=v1[i][j]) {
                    break;
                }
                st=v[pos].second;
            } 
            cnt+=(j>=v1[i].length());
        }
        return cnt;

    }
};
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