# 916. Word Subsets 单词子集

## # 题目描述：

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"]
``````

Example 2:

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

Example 3:

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

Example 4:

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

Example 5:

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

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]`.

## # 解题方法

``````class Solution(object):
def wordSubsets(self, A, B):
"""
:type A: List[str]
:type B: List[str]
:rtype: List[str]
"""
B = set(B)
res = []
count = collections.defaultdict(int)
for b in B:
cb = collections.Counter(b)
for c, v in cb.items():
count[c] = max(count[c], v)
res = []
for a in A:
ca = collections.Counter(a)
isSuccess = True
for c, v in count.items():
if v > ca[c]:
isSuccess = False
break
if isSuccess:
res.append(a)
return res
``````

https://leetcode.com/problems/word-subsets/discuss/175854/C++JavaPython-Straight-Forward

## # 日期

2018 年 9 月 30 日 —— 9月最后一天啦！