916. Word Subsets - Medium

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].

 

when we check whether wordA in A is a superset of wordB, we are individually checking the counts of letters: for each letter, N_letter(wordA) >= N_letter(wordB). this is the same as checking N_letter(wordA) >= max (N_letter(wordB))

in other words, reduce B to a single word bmax, then compare the counts of letters between words a in A, and bmax

time = O(A + B), A and B is the total amount of information in A and B respectively, space = O(A.length + B.length)

 

class Solution {
    public List<String> wordSubsets(String[] A, String[] B) {
        int[] bmax = new int[26];
        
        for(String b : B) {
            int[] bCount = count(b);
            for(int i = 0; i < 26; i++) {
                bmax[i] = Math.max(bmax[i], bCount[i]);
            }
        }
        
        List<String> res = new ArrayList<>();
        for(String a : A) {
            int[] aCount = count(a);
            for(int i = 0; i < 26; i++) {
                if(aCount[i] < bmax[i]) {
                    break;
                }
                if(i == 25) {
                    res.add(a);
                }
            }
        }
        return res;
    }
    
    public int[] count(String s) {
        int[] counter = new int[26];
        for(char c : s.toCharArray()) {
            counter[c - 'a']++;
        }
        return counter;
    }
}