Formatted question description: https://leetcode.ca/all/916.html

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].

Related Topics:
String

Solution 1.

// OJ: https://leetcode.com/problems/word-subsets/

// Time: O((A + B) * N) where N is the maximum length of string
// Space: O(1)
class Solution {
public:
    vector<string> wordSubsets(vector<string>& A, vector<string>& B) {
        int N = A.size(), target[26] = {}, j = 0;
        for (auto &b : B) {
            int cnt[26] = {};
            for (char c : b) cnt[c - 'a']++;
            for (int i = 0; i < 26; ++i) target[i] = max(target[i], cnt[i]);
        }
        for (int i = 0; i < N; ++i) {
            int cnt[26] = {};
            for (char c : A[i]) cnt[c - 'a']++;
            bool valid = true;
            for (int k = 0; k < 26 && valid; ++k) valid = target[k] <= cnt[k];
            if (valid) A[j++] = A[i];
        }
        A.resize(j);
        return A;
    }
};

Java

class Solution {
    public List<String> wordSubsets(String[] A, String[] B) {
        List<String> universalList = new ArrayList<String>();
        int[] counts = getMaxCounts(B);
        for (String word : A) {
            int[] wordCounts = getWordCounts(word);
            boolean flag = true;
            for (int i = 0; i < 26; i++) {
                if (wordCounts[i] < counts[i]) {
                    flag = false;
                    break;
                }
            }
            if (flag)
                universalList.add(word);
        }
        return universalList;
    }

    public int[] getMaxCounts(String[] array) {
        int[] counts = new int[26];
        for (String word : array) {
            int[] curCounts = new int[26];
            char[] wordArray = word.toCharArray();
            for (char c : wordArray)
                curCounts[c - 'a']++;
            for (int i = 0; i < 26; i++)
                counts[i] = Math.max(counts[i], curCounts[i]);
        }
        return counts;
    }

    public int[] getWordCounts(String word) {
        int[] wordCounts = new int[26];
        char[] wordArray = word.toCharArray();
        for (char c : wordArray)
            wordCounts[c - 'a']++;
        return wordCounts;
    }
}

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