916. Word Subsets

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916. Word Subsets

Description

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:

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

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

Example 2:

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

Output: ["leetcode"]

Example 3:

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

Output: ["cccbb"]

Constraints:

  • 1 <= words1.length, words2.length <= 10^4
  • 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 .

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    vector<string> wordSubsets(vector<string>& words1, vector<string>& words2) {
        vector<int> targets(26, 0);
        for (string& word : words2) {
            vector<int> curWord(26, 0);
            for (char& c : word) {
                curWord[c - 'a']++;
                targets[c - 'a'] = max(targets[c - 'a'], curWord[c - 'a']);
            }
        }
        vector<string> res;
        for (string& word : words1) {
            vector<int> curWord(26, 0);
            for (char& c : word) {
                curWord[c - 'a']++;
            }
            bool valid_word = true;
            for (int i = 0; i < targets.size(); i++) {
                if (targets[i] > curWord[i]) {
                    valid_word = false;
                    break;
                }
            }
            if (valid_word) {
                res.push_back(word);
            }
        }
        return res;
    }
};