Count Common Words With One Occurrence

Given two string arrays words1 and words2, return the number of strings that appear exactly once in each of the two arrays.

Example 1:

Input: words1 = ["leetcode","is","amazing","as","is"], words2 = ["amazing","leetcode","is"]
Output: 2
Explanation:

• "leetcode" appears exactly once in each of the two arrays. We count this string.
• "amazing" appears exactly once in each of the two arrays. We count this string.
• "is" appears in each of the two arrays, but there are 2 occurrences of it in words1. We do not count this string.
• "as" appears once in words1, but does not appear in words2. We do not count this string. Thus, there are 2 strings that appear exactly once in each of the two arrays.

Example 2:

Input: words1 = ["b","bb","bbb"], words2 = ["a","aa","aaa"]
Output: 0
Explanation: There are no strings that appear in each of the two arrays.

Example 3:

Input: words1 = ["a","ab"], words2 = ["a","a","a","ab"]
Output: 1
Explanation: The only string that appears exactly once in each of the two arrays is "ab".

Constraints:

• 1 <= words1.length, words2.length <= 1000
• 1 <= words1[i].length, words2[j].length <= 30
• words1[i] and words2[j] consists only of lowercase English letters.

SOLUTION:

from collections import Counter

class Solution:
    def countWords(self, words1: List[str], words2: List[str]) -> int:
        w1 = Counter(words1)
        w2 = Counter(words2)
        op = set()
        for a in w1:
            if w1[a] == 1 and w2[a] == 1:
                op.add(a)
        for a in w2:
            if w1[a] == 1 and w2[a] == 1:
                op.add(a)
        return len(op)