With respect to a given puzzle string, a word is valid if both the following conditions are satisfied:

• word contains the first letter of puzzle.
• For each letter in word, that letter is in puzzle.
  For example, if the puzzle is "abcdefg", then valid words are "faced", "cabbage", and "baggage"; while invalid words are "beefed" (doesn't include "a") and "based" (includes "s" which isn't in the puzzle).

Return an array answer, where answer[i] is the number of words in the given word list words that are valid with respect to the puzzle puzzles[i].

 

Example :

Input: 
words = ["aaaa","asas","able","ability","actt","actor","access"], 
puzzles = ["aboveyz","abrodyz","abslute","absoryz","actresz","gaswxyz"]
Output: [1,1,3,2,4,0]
Explanation:
1 valid word for "aboveyz" : "aaaa" 
1 valid word for "abrodyz" : "aaaa"
3 valid words for "abslute" : "aaaa", "asas", "able"
2 valid words for "absoryz" : "aaaa", "asas"
4 valid words for "actresz" : "aaaa", "asas", "actt", "access"
There're no valid words for "gaswxyz" cause none of the words in the list contains letter 'g'.

 

Constraints:

• 1 <= words.length <= 10^5
• 4 <= words[i].length <= 50
• 1 <= puzzles.length <= 10^4
• puzzles[i].length == 7
• words[i][j], puzzles[i][j] are English lowercase letters.
• Each puzzles[i] doesn't contain repeated characters.

思路：关键是抓住puzzle的长度是7，直接枚举即可，去掉第一位，6位组合最多2**6=64

import itertools
from collections import Counter
class Solution(object):
    def findNumOfValidWords(self, words, puzzles):
        """
        :type words: List[str]
        :type puzzles: List[str]
        :rtype: List[int]
        """
        cnt = Counter(''.join(sorted(set(w))) for w in words)
        res = [0]*len(puzzles)
        for idx,p in enumerate(puzzles):
            first = p[0]
            t = set(p[1:])
            for i in range(len(t)+1):
                for cs in itertools.combinations(t, i):
                    cs = set(list(cs)+[first])
                    res[idx] += cnt.get(''.join(sorted(list(cs))), 0)
        return res

Trie也是可以的，参考https://leetcode.com/problems/number-of-valid-words-for-each-puzzle/discuss/371944/Python-Trie-O(km%2Bn)