Solve in Python:
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 <= A.length, B.length <= 10000
1 <= A[i].length, B[i].length <= 10
A[i] and B[i] consist only of lowercase letters.
All words in A[i] are unique: there isn't i != j with A[i] == A[j].
class Solution:
    def wordSubsets(self, A: List[str], B: List[str]) -> List[str]:
        \"\"\"def checkSubset(a, B):
            aDict = {}
            bDict = {}
            for b in B:
                bSet = set(b)
                for c in bSet:
                    if b.count(c) > a.count(c):
                        return False
            return True
        ret = []
        for a in A:
            if not checkSubset(a, B):
                continue
            else:
                ret.append(a)
        return ret\"\"\"
        \"\"\"def convertB(b):
            d = {}
            for c in b: 
                if c not in d:
                    d[c] = 1
                else:
                    d[c]+=1
            return d 
        for i in range(len(B)):
            B[i] = convertB(B[i])
        def convertA(a):
            d = {}
            for c in a:
                if c not in d:
                    d[c]=1
                else:
                    d[c]+=1
            return d
        convertedA = []
        for i in range(len(A)):
            convertedA.append(convertA(A[i]))
        
        def isGoodWord(aDict):
            for wordDict in B:
                for c in wordDict:
                    if c not in aDict or wordDict[c] > aDict[c]:
                        return False
            return True
        ret = []
        for i in range(len(convertedA)):
            if isGoodWord(convertedA[i]):
                ret.append(A[i])
        return ret\"\"\"
        bDict = {}
        def convertB(b):
            d = {}
            for c in b: 
                if c not in d:
                    d[c] = 1
                else:
                    d[c]+=1
            return d 
        for i in range(len(B)):
            B[i] = convertB(B[i])
        for d in B:
            for key in d:
                if key not in bDict:
                    bDict[key] = d[key]
                else:
                    if bDict[key] < d[key]:
                        bDict[key] = d[key]
        def convertA(a):
    ...