Create 916._Word_Subsets.md

Author: 0xkookoo

docs/Leetcode_Solutions/Python/916._Word_Subsets.md

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+# 916. Word Subsets
+
+**<font color=red>难度: Medium</font>**
+
+## 刷题内容
+
+> 原题连接
+
+* https://leetcode.com/contest/weekly-contest-104/problems/word-subsets/
+
+> 内容描述
+
+```
+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"]
+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].
+```
+
+## 解题方案
+
+> 思路 1
+******- 时间复杂度: O(max(ma*la, mb*lb))******- 空间复杂度: O(1)******
+
+
+假设A的长度为na，其中最长的一个单词长度为la
+B的长度为nb，其中最长的一个单词长度为lb
+
+那么时间复杂度为O(max(ma*la, mb*lb))，由于字符总共就26个，所以空间复杂度可以看作O(1)
+
+
+先求出B中所有字符在所有单词中出现的最大次数，用lookup记录，然后如果A中的某一个单词满足lookup里面所有的key都有，且对应数量都大于等于lookup中的，
+那么该单词满足条件，append到最终结果中
+
+```python
+class Solution(object):
+    def wordSubsets(self, A, B):
+        """
+        :type A: List[str]
+        :type B: List[str]
+        :rtype: List[str]
+        """
+        if not B or len(B) == 0:
+            return A
+        lookup = collections.Counter(B[0])
+        for word in B[1:]:
+            tmp = collections.Counter(word)
+            for key in tmp.keys():
+                if key not in lookup:
+                    lookup[key] = tmp[key]
+                else:
+                    lookup[key] = max(lookup[key], tmp[key])
+        def uni(tmp_a, tmp_b):
+            if len(tmp_a.keys()) < len(tmp_b.keys()):
+                return False
+            for key in tmp_b.keys():
+                if key not in tmp_a:
+                    return False
+                else:
+                    if tmp_a[key] < tmp_b[key]:
+                        return False
+            return True
+        res = []
+        for word in A:
+            if uni(collections.Counter(word), lookup):
+                res.append(word)
+        return res
+```
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