【leetcode】1178. Number of Valid Words for Each Puzzle

题目如下：

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.

解题思路：本题可以采用逆向思维法。首先把words中的每个单词都做去掉重复字符处理，然后按字典序重新把字符排序，得到新的单词，例如 ealebb 经过处理后就是 able。接下来对puzzles中的每个单词除第一个字符外的子字符串做同样的处理，例如 gawbxyz 处理后的结果是 g + abwxyz 两部分。对于满足gawbxyz谜底的单词要具备的条件是 第一个字符g是必定存在，同时后面的abwxyz 任意选择0~6个字符，这要判断这些组合有哪些在处理words中出现过，即可得puzzles中每个puzzle的结果。因为puzzles[i].length == 7，所以最多只会有2^6次方中组合，不会存在性能问题。

代码如下：

class Solution(object):
    def findNumOfValidWords(self, words, puzzles):
        """
        :type words: List[str]
        :type puzzles: List[str]
        :rtype: List[int]
        """
        dic_words = {}
        for word in words:
            newword = ''.join(sorted(list(set(list(word)))))
            dic_words[newword] = dic_words.setdefault(newword,0) + 1
        res = []
        for puzzle in puzzles:
            count = 0
            first = puzzle[0]
            puzzle = ''.join(sorted(list(set(list(puzzle[1:])))))
            if puzzle.find(first) != -1:
                inx = puzzle.find(first)
                puzzle = puzzle[:inx] + puzzle[inx+1:]
            if first in dic_words:
                count += dic_words[first]
            queue = []
            for (i,v) in enumerate(puzzle):
                queue.append((v,i))
            while len(queue) > 0:
                v,i = queue.pop(0)
                if i == len(puzzle) - 1:
                    key = ''.join(sorted(list(first + v)))
                    if key in dic_words:
                        count += dic_words[key]
                    continue
                queue.append((v,i+1))
                queue.append((v + puzzle[i+1],i+1))
            res.append(count)
        return res