916. Word Subsets

Problem Link

Description

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You are given two string arrays words1 and words2.

A string b is a subset of string 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".

A string a from words1 is universal if for every string b in words2, b is a subset of a.

Return an array of all the universal strings in words1. You may return the answer in any order.

 

Example 1:

Input: words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["e","o"]

Output: ["facebook","google","leetcode"]

Example 2:

Input: words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["lc","eo"]

Output: ["leetcode"]

Example 3:

Input: words1 = ["acaac","cccbb","aacbb","caacc","bcbbb"], words2 = ["c","cc","b"]

Output: ["cccbb"]

 

Constraints:

• 1 <= words1.length, words2.length <= 104
• 1 <= words1[i].length, words2[i].length <= 10
• words1[i] and words2[i] consist only of lowercase English letters.
• All the strings of words1 are unique.

Solution

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Python3

class Solution:
    def wordSubsets(self, words1: List[str], words2: List[str]) -> List[str]:
        res = []
        counter = [0] * 26
 
        for word in words2:
            count = [0] * 26
 
            for c in word:
                count[ord(c) - ord('a')] += 1
            
            for i in range(26):
                counter[i] = max(counter[i], count[i])
 
        for word in words1:
            count = [0] * 26
 
            for c in word:
                count[ord(c) - ord('a')] += 1
            
            ok = True
            for i in range(26):
                if counter[i] > count[i]:
                    ok = False
                    break
            
            if ok:
                res.append(word)
 
        return res