word subsets medium

Author: weiy

Array/WordSubsets.py

@@ -0,0 +1,86 @@
+"""
+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].
+
+
+如果 B 中每一个元素出现的字符均在 A 中某一个元素中出现，则视为子单词，给一个A 和 B求子单词数量。
+
+思路：
+哈希 A,
+哈希 B。
+
+B 哈希的时候要注意，有重复。
+
+比如["ec","oc","ceo"]
+只要有 c e o 各一次即可。不需要每个都判断一次。
+
+测试地址：
+https://leetcode.com/problems/word-subsets/description/
+
+这个写法比较慢，前面的思路基本都是哈希。
+
+"""
+from collections import Counter
+
+class Solution(object):
+    def wordSubsets(self, A, B):
+        """
+        :type A: List[str]
+        :type B: List[str]
+        :rtype: List[str]
+        """
+        dict_A = {i:Counter(i) for i in A}
+        dict_B = {}
+        
+        for i in B:
+            c = Counter(i)
+            for j in c:
+                if c.get(j) > dict_B.get(j, 0):
+                    dict_B[j] = c.get(j)
+                    
+        result = []
+        
+        for i in dict_A:
+            for j in dict_B:
+                if dict_B[j] > dict_A[i].get(j):
+                    break
+            else:
+                result.append(i)
+                    
+        return result