<p>Given a list of <code>words</code>, list of&nbsp; single&nbsp;<code>letters</code> (might be repeating)&nbsp;and <code>score</code>&nbsp;of every character.</p>

<p>Return the maximum score of <strong>any</strong> valid set of words formed by using the given letters (<code>words[i]</code> cannot be used two&nbsp;or more times).</p>

<p>It is not necessary to use all characters in <code>letters</code> and each letter can only be used once. Score of letters&nbsp;<code>&#39;a&#39;</code>, <code>&#39;b&#39;</code>, <code>&#39;c&#39;</code>, ... ,<code>&#39;z&#39;</code> is given by&nbsp;<code>score[0]</code>, <code>score[1]</code>, ... , <code>score[25]</code> respectively.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;dog&quot;,&quot;cat&quot;,&quot;dad&quot;,&quot;good&quot;], letters = [&quot;a&quot;,&quot;a&quot;,&quot;c&quot;,&quot;d&quot;,&quot;d&quot;,&quot;d&quot;,&quot;g&quot;,&quot;o&quot;,&quot;o&quot;], score = [1,0,9,5,0,0,3,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0]
<strong>Output:</strong> 23
<strong>Explanation:</strong>
Score  a=1, c=9, d=5, g=3, o=2
Given letters, we can form the words &quot;dad&quot; (5+1+5) and &quot;good&quot; (3+2+2+5) with a score of 23.
Words &quot;dad&quot; and &quot;dog&quot; only get a score of 21.</pre>

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

<pre>
<strong>Input:</strong> words = [&quot;xxxz&quot;,&quot;ax&quot;,&quot;bx&quot;,&quot;cx&quot;], letters = [&quot;z&quot;,&quot;a&quot;,&quot;b&quot;,&quot;c&quot;,&quot;x&quot;,&quot;x&quot;,&quot;x&quot;], score = [4,4,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,5,0,10]
<strong>Output:</strong> 27
<strong>Explanation:</strong>
Score  a=4, b=4, c=4, x=5, z=10
Given letters, we can form the words &quot;ax&quot; (4+5), &quot;bx&quot; (4+5) and &quot;cx&quot; (4+5) with a score of 27.
Word &quot;xxxz&quot; only get a score of 25.</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;leetcode&quot;], letters = [&quot;l&quot;,&quot;e&quot;,&quot;t&quot;,&quot;c&quot;,&quot;o&quot;,&quot;d&quot;], score = [0,0,1,1,1,0,0,0,0,0,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,0]
<strong>Output:</strong> 0
<strong>Explanation:</strong>
Letter &quot;e&quot; can only be used once.</pre>

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

<ul>
	<li><code>1 &lt;= words.length &lt;= 14</code></li>
	<li><code>1 &lt;= words[i].length &lt;= 15</code></li>
	<li><code>1 &lt;= letters.length &lt;= 100</code></li>
	<li><code>letters[i].length == 1</code></li>
	<li><code>score.length ==&nbsp;26</code></li>
	<li><code>0 &lt;= score[i] &lt;= 10</code></li>
	<li><code>words[i]</code>, <code>letters[i]</code>&nbsp;contains only lower case English letters.</li>
</ul>

class Solution:
    def maxScoreWords(self, words: List[str], letters: List[str], score: List[int]) -> int:
        
        # list of words. 
        # list of letters. might be repeating 
        
        # scores of a-z are given 

        # words[i] cannot be used twice
        # each letter should be used only once 

        # which should i choose ? 
        # d + o + g 
        # g + o + o + d 
        # if i have enough characters, choose "good"

        # maximum score of any valid set of words 
        # more length -> more score ? 
        # more length + better score -> more score ? 
        
        # get scores of individual words
        # score if possible else 0 

        # what combination gives the maximum score ? 
        # but the problem is, if i choose one word, i cannot chooose another 
        # given the constraints i think this is the approach 

        counter = Counter(letters)
        wscores = [0] * len(words)
        for i, word in enumerate(words):
            wcounter = Counter(word)
            s = 0 
            for k in wcounter:
                if wcounter[k] <= counter[k]:
                    s += wcounter[k] * score[ord(k)-ord('a')]
                else:
                    s = 0
                    break
            wscores[i] = s

        def possible_to_take(index, counter):
            wc = Counter(words[index])
            for k in wc:
                if wc[k] > counter[k]:
                    return False
            return True

        def recur(index):
            if index >= len(words): return 0
            take, notake = 0, 0
            if possible_to_take(index, counter):
                for k in words[index]: counter[k] -= 1
                take = wscores[index] + recur(index+1)
                for k in words[index]: counter[k] += 1
            notake = recur(index+1)
            return max(take, notake)
        return recur(0)

16700