count-common-words-with-one-occurrence_2085.py

# 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".

# ---------------------------------------Runtime 83 ms Beats 74.70% Memory 16.7 MB Beats 74.15%---------------------------------------
from collections import Counter

from typing import List


# My solution
# Time Complexity O(n)
class Solution:
    def countWords(self, words1: List[str], words2: List[str]) -> int:
        w1: Counter = Counter(words1)
        w2: Counter = Counter(words2)
        compare: list = [x for x in w1 if w1.get(x) == 1]
        compare.extend([x for x in w1 if w2.get(x) == 1])
        out: Counter = Counter(compare)
        return len([x for x in out if out.get(x) == 2])