With respect to a given <code>puzzle</code> string, a <code>word</code> is <em>valid</em> if both the following conditions are satisfied:
<ul>
	<li><code>word</code> contains the first letter of <code>puzzle</code>.</li>
	<li>For each letter in <code>word</code>, that letter is in <code>puzzle</code>.
	<ul>
		<li>For example, if the puzzle is <code>&quot;abcdefg&quot;</code>, then valid words are <code>&quot;faced&quot;</code>, <code>&quot;cabbage&quot;</code>, and <code>&quot;baggage&quot;</code>, while</li>
		<li>invalid words are <code>&quot;beefed&quot;</code> (does not include <code>&#39;a&#39;</code>) and <code>&quot;based&quot;</code> (includes <code>&#39;s&#39;</code> which is not in the puzzle).</li>
	</ul>
	</li>
</ul>
Return <em>an array </em><code>answer</code><em>, where </em><code>answer[i]</code><em> is the number of words in the given word list </em><code>words</code><em> that is valid with respect to the puzzle </em><code>puzzles[i]</code>.
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;aaaa&quot;,&quot;asas&quot;,&quot;able&quot;,&quot;ability&quot;,&quot;actt&quot;,&quot;actor&quot;,&quot;access&quot;], puzzles = [&quot;aboveyz&quot;,&quot;abrodyz&quot;,&quot;abslute&quot;,&quot;absoryz&quot;,&quot;actresz&quot;,&quot;gaswxyz&quot;]
<strong>Output:</strong> [1,1,3,2,4,0]
<strong>Explanation:</strong> 
1 valid word for &quot;aboveyz&quot; : &quot;aaaa&quot; 
1 valid word for &quot;abrodyz&quot; : &quot;aaaa&quot;
3 valid words for &quot;abslute&quot; : &quot;aaaa&quot;, &quot;asas&quot;, &quot;able&quot;
2 valid words for &quot;absoryz&quot; : &quot;aaaa&quot;, &quot;asas&quot;
4 valid words for &quot;actresz&quot; : &quot;aaaa&quot;, &quot;asas&quot;, &quot;actt&quot;, &quot;access&quot;
There are no valid words for &quot;gaswxyz&quot; cause none of the words in the list contains letter &#39;g&#39;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;apple&quot;,&quot;pleas&quot;,&quot;please&quot;], puzzles = [&quot;aelwxyz&quot;,&quot;aelpxyz&quot;,&quot;aelpsxy&quot;,&quot;saelpxy&quot;,&quot;xaelpsy&quot;]
<strong>Output:</strong> [0,1,3,2,0]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= words.length &lt;= 10<sup>5</sup></code></li>
	<li><code>4 &lt;= words[i].length &lt;= 50</code></li>
	<li><code>1 &lt;= puzzles.length &lt;= 10<sup>4</sup></code></li>
	<li><code>puzzles[i].length == 7</code></li>
	<li><code>words[i]</code> and <code>puzzles[i]</code> consist of lowercase English letters.</li>
	<li>Each <code>puzzles[i] </code>does not contain repeated characters.</li>
</ul>

class Solution {
public:
    vector<int> findNumOfValidWords(vector<string>& words, vector<string>& puzzles) {
        int SIZE = 26;                                   // total character size
        vector<vector<int>> trie = {vector<int>(SIZE)};  // we use vector to mimic the trie tree
        vector<int> count = {0};                         // the number of words ending at node i
        for (string word : words) {  // we do not use reference here to avoid modifying the input
            sort(word.begin(), word.end());
            word.erase(unique(word.begin(), word.end()), word.end());
            if (word.size() <= 7) {  // longer words are never valid
                // insert into trie
                int node = 0;
                for (char& letter : word) {
                    int i = letter - 'a';
                    if (trie[node][i] == 0) {  // push empty node
                        trie.push_back(vector<int>(SIZE));
                        count.push_back(0);
                        trie[node][i] = trie.size() - 1;
                    }
                    node = trie[node][i];
                }
                count[node]++;
            }
        }
        // search for valid words
        function<int(int, bool, string&)> dfs = [&](int node, bool hasFirst, string& puzzle) {
            int total = hasFirst ? count[node] : 0;
            for (char& letter : puzzle) {
                int i = letter - 'a';
                if (trie[node][i]) {
                    total += dfs(trie[node][i], hasFirst || (letter == puzzle[0]), puzzle);
                }
            }
            return total;
        };
        vector<int> result;
        for (string& puzzle : puzzles) {
            result.push_back(dfs(0, false, puzzle));
        }
        return result;
    }
};
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