【LeetCode】916. Word Subsets 解题报告（Python）

作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

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题目地址: https://leetcode.com/problems/word-subsets/description/

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"]
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].

题目大意

如果B中的每一个元素中的每个字符都能在A中的某个字符串中找到，那么这个是个符合要求的字符串。求A中所有满足要求的字符串。

解题方法

如果按照题目要求的意思去直接做，那么要遍历B的每个元素的每个字符与A对应，这个时间复杂度肯定过不了OJ。所以，采用了一个很巧妙的方法，把B当做一个限制条件，直接求解这个限制条件。

对B的每个元素遍历，然后统计每个元素中每个字符串出现的次数，更新整体限制条件为每个元素在字符串中出现的次数的最大值。

统计结束之后，对A遍历的时候，只要看A是否满足这个限制条件就行了，所以挺快的。

时间复杂度是O(N)，空间复杂度是O(N)。

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

参考资料：

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！