You are given two string arrays words1 and words2. A string w in words1 is called a universal word if, for every string b in words2, b is a subset of w in terms of character counts. In other words, for each lowercase letter, w must contain that letter at least as many times as it appears in b.

Return all strings in words1 that are universal with respect to words2.

Constraints:

• $1 \leq$ words1.length, words2.length $\leq 10^4$

• $1 \leq$ words1[i].length, words2[i].length $\leq 10$

• words1[i] and words2[i] consist only of lowercase English letters

• All strings in words1 are unique

Input: words1 = ["amazon","apple","facebook","google","leetcode"]-List[str], words2 = ["e","o"]-List[str]
Output: ["facebook","google","leetcode"]-List[str]
Explanation: To be universal, a word must contain at least one e and at least one o (the maximum required counts across words2). "facebook", "google", and "leetcode" have both and qualify, while "amazon" lacks e and "apple" lacks o.

Input: words1 = ["amazon","apple","facebook","google","leetcode"]-List[str], words2 = ["l","e"]-List[str]
Output: ["apple","google","leetcode"]-List[str]
Explanation: A universal word must contain at least one l and at least one e. "apple", "google", and "leetcode" have both and qualify, while "amazon" lacks both l and e and "facebook" lacks l.

Input: words1 = ["amazon","apple","facebook","google","leetcode"]-List[str], words2 = ["e","oo"]-List[str]
Output: ["facebook","google"]-List[str]
Explanation: Taking the maximum frequency per letter across words2, "e" requires e:1 and "oo" requires o:2, so every universal word needs at least one e and at least two os. "facebook" and "google" (e:1, o:2) qualify, while "leetcode" has only one o, "amazon" has one o and no e, and "apple" has no o.

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