916. 单词子集

给你两个字符串数组 words1 和 words2。

现在，如果 b 中的每个字母都出现在 a 中，包括重复出现的字母，那么称字符串 b 是字符串 a 的 子集 。

例如，"wrr" 是 "warrior" 的子集，但不是 "world" 的子集。
如果对 words2 中的每一个单词 b，b 都是 a 的子集，那么我们称 words1 中的单词 a 是 通用单词 。

以数组形式返回 words1 中所有的通用单词。你可以按 任意顺序 返回答案。

示例 1：

输入：words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["e","o"]
输出：["facebook","google","leetcode"]
示例 2：

输入：words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["l","e"]
输出：["apple","google","leetcode"]
示例 3：

输入：words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["e","oo"]
输出：["facebook","google"]
示例 4：

输入：words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["lo","eo"]
输出：["google","leetcode"]
示例 5：

输入：words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["ec","oc","ceo"]
输出：["facebook","leetcode"]
 

提示：

1 <= words1.length, words2.length <= 104
1 <= words1[i].length, words2[i].length <= 10
words1[i] 和 words2[i] 仅由小写英文字母组成
words1 中的所有字符串 互不相同

class Solution:
    def wordSubsets(self, words1: List[str], words2: List[str]) -> List[str]:
        # 1. 首先根据words2得到每个单词中每个字母的最大个数
        # 2. 和words1的每个字符串进行比较

        bMax = [0] * 26
        for word2 in words2:
            for i, b in enumerate(self.helper(word2)):
                bMax[i] = max(b, bMax[i]) # 得到每个字母的最大个数, 非总数

        res = []
        for word1 in words1:
            if all([x >= y for x, y in zip(self.helper(word1), bMax)]):
                res.append(word1)
        
        return res

    # 计数
    def helper(self, word):
        _c = [0] * 26

        for s in word:
            _c[ord(s) - ord('a')] += 1

        return _c