leetcode Word Subsets

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本文介绍了一种算法，用于从两个字符串数组中找出所有包含B数组中每个元素子集的A数组中的单词。通过实例展示了如何使用Counter类来比较字符频率，确保A数组中的单词能覆盖B数组中所有单词的字符需求。

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].

Accepted

15,865

Submissions

34,051

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TLE code without merging required characters:

from collections import Counter

class Solution:
    def a_contains_b(self, a, b):
        for k,v in b.items():
            if (k not in a or (k in a and a[k] < v)):
                return False
        return True

    def contains_all(self, a, b_list):
        for b in b_list:
            is_contain = self.a_contains_b(a,b)
            if (is_contain == False):
                return False
        return True

    def wordSubsets(self, A, B):
        b_list = []
        for b in B:
            b_list.append(Counter(b))
        res = []
        for a in A:
            c_a = Counter(a)
            if (self.contains_all(c_a, b_list)):
                res.append(a)
        return res

s = Solution()
print(s.wordSubsets(["amazon","apple","facebook","google","leetcode"],["l","e"]))

Merge required characters:

from collections import Counter

class Solution:
    def a_contains_b(self, a, b):
        for k,v in b.items():
            if (k not in a or (k in a and a[k] < v)):
                return False
        return True

    def wordSubsets(self, A, B):
        required_letter = Counter()
        for b in B:
            b_c = Counter(b)
            for k,v in b_c.items():
                if (k not in required_letter or (k in required_letter and required_letter[k] < v)):
                    required_letter[k] = v
        res = []
        for a in A:
            c_a = Counter(a)
            if (self.a_contains_b(c_a, required_letter)):
                res.append(a)
        return res

s = Solution()
print(s.wordSubsets(["amazon","apple","facebook","google","leetcode"],["e","o"]))