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].

Solution: Hashtable

Find the max requirement for each letter from B.

Time complexity: O(|A|+|B|)

Space complexity: O(26)

// Author: Huahua, 104 ms
class Solution {
public:
  vector<string> wordSubsets(vector<string>& A, vector<string>& B) {
    vector<int> req(26);
    for (const string& b : B) {
      vector<int> cur(getCount(b));
      for (int i = 0; i < 26; ++i)
        req[i] = max(req[i], cur[i]);
    }
    
    vector<string> ans;    
    for (const string& a : A) {
      vector<int> cur(getCount(a));
      bool universal = true;
      for (int i = 0; i < 26; ++i)
        if (cur[i] < req[i]) {
          universal = false;
          break; 
        }
      if (universal) ans.push_back(a);
    }
    return ans;
  }
private:
  vector<int> getCount(const string& a) {
    vector<int> count(26);
    for (char c : a)
      ++count[c - 'a'];
    return count;
  }
};

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