Description

We are given two arrays A and B of words. Each word is a string of lowercase letters.

Now, say that word b is a subset of word a if every letter in b occurs in a, including multiplicity. For example, “wrr” is a subset of “warrior”, but is not a subset of “world”.

Now say a word a from A is universal if for every b in B, b is a subset of a.

Return a list of all universal words in A. You can return the words in any order.

Example 1:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["e","o"]
Output: ["facebook","google","leetcode"]

Example 2:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["l","e"]
Output: ["apple","google","leetcode"]

Example 3:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["e","oo"]
Output: ["facebook","google"]

Example 4:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["lo","eo"]
Output: ["google","leetcode"]

Example 5:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["ec","oc","ceo"]
Output: ["facebook","leetcode"]

Note:

  1. 1 <= A.length, B.length <= 10000
  2. 1 <= A[i].length, B[i].length <= 10
  3. A[i] and B[i] consist only of lowercase letters.
  4. All words in A[i] are unique: there isn’t i != j with A[i] == A[j].

分析

题目的意思是：给定一个字符串数组A，判断b中所有的字符是否是a的子集，我一开始用字典做了一下，发现b中的字符是字符串，然后没有ac。我参考了一下答案的思路，count统计的是word的词频，bmax统计的是b中每个字符串最大词频。如果A能够满足b对于字符的最大词频，则满足条件，如果知道这个，剩下的就是遍历A看A中的字符串能够满足条件了哈。

代码

class Solution:
    def count(self,word):
        ans=[0]*26
        for ch in word:
            ans[ord(ch)-ord('a')]+=1
        return ans
    
    def wordSubsets(self, A: List[str], B: List[str]) -> List[str]:
        bmax=[0]*26
        for b in B:
            for i,c in enumerate(self.count(b)):
                bmax[i]=max(bmax[i],c)
                
        res=[]
        for a in A:
            flag=True
            for x,y in zip(self.count(a),bmax):
                if(x<y):
                    flag=False
                    break
            if(flag):
                res.append(a)
        return res

参考文献

​​[LeetCode] solution​​