916. Word Subsets

Problem

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 = ["l","e"]
Output: ["apple","google","leetcode"]

  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.

Solution (Java)

class Solution {
    public List<String> wordSubsets(String[] words1, String[] words2) {
        List<String> l1 = new ArrayList<>();
        int[] target = new int[26];
        for (String s1 : words2) {
            int[] temp = new int[26];
            for (char ch1 : s1.toCharArray()) {
                temp[ch1 - 'a']++;
                target[ch1 - 'a'] = Math.max(target[ch1 - 'a'], temp[ch1 - 'a']);
            }
        }
        for (String s1 : words1) {
            int[] count = new int[26];
            for (char ch1 : s1.toCharArray()) {
                count[ch1 - 'a']++;
            }

            if (checkIt(target, count)) {
                l1.add(s1);
            }
        }
        return l1;
    }

    private boolean checkIt(int[] target, int[] count) {
        for (int i = 0; i < 26; i++) {
            if (count[i] < target[i]) {
                return false;
            }
        }
        return true;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).