<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 Node {
public:
    vector<Node*> nodes;
    int isEnd;

    Node() {
        nodes.resize(26,NULL);
        isEnd = 0;
    }
};

class Solution {
public:
    int numMatchingSubseq(string s, vector<string>& words) {
        Node* tmp = new Node();

        for(int c1=0;c1<words.size();c1++) {
            rec(tmp,0,words[c1]);
        }

        int ans = 0;
        vector<queue<Node*>> alp(26);

        for(int c1=0;c1<26;c1++) {
            if(tmp->nodes[c1]) {
                alp[c1].push(tmp->nodes[c1]);
            }
        }

        for(int c1=0;c1<s.length();c1++) {
            int ind = s[c1]-'a';
            int sz = alp[ind].size();

            while(sz--) {
                Node* curr = alp[ind].front();
                alp[ind].pop();
                ans += curr->isEnd;
                for(int c2=0;c2<26;c2++) {
                    if(curr->nodes[c2]) {
                        alp[c2].push(curr->nodes[c2]);
                    }
                }
            }
        }

        return ans;
    }

    void rec(Node* curr, int ind, string& s) {
        if(ind==s.length()) {
            curr->isEnd += 1;
            return;
        }

        if(curr->nodes[s[ind]-'a']==NULL)
            curr->nodes[s[ind]-'a'] = new Node();

        rec(curr->nodes[s[ind]-'a'],ind+1,s);

        return;
    }
};
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