The statistics of leetcode simple question is the public string that has appeared once

subject

Here are two string arrays words1 and words2 , Please return in two string arrays Just once Number of strings .
Example 1：
Input ：words1 = [“leetcode”,“is”,“amazing”,“as”,“is”], words2 = [“amazing”,“leetcode”,“is”]
Output ：2
explain ：

• “leetcode” Occurs exactly once in both arrays , Count answers .
• “amazing” Occurs exactly once in both arrays , Count answers .
• “is” In both arrays , But in words1 In the 2 Time , Don't count the answers .
• “as” stay words1 There's a time when , But in words2 There's no such thing as , Don't count the answers .
  therefore , Yes 2 A string appears exactly once in both arrays .
  Example 2：
  Input ：words1 = [“b”,“bb”,“bbb”], words2 = [“a”,“aa”,“aaa”]
  Output ：0
  explain ： No string occurs exactly once in both arrays .
  Example 3：
  Input ：words1 = [“a”,“ab”], words2 = [“a”,“a”,“a”,“ab”]
  Output ：1
  explain ： The only string that appears once in both arrays is “ab” .
  Tips ：
  1 <= words1.length, words2.length <= 1000
  1 <= words1[i].length, words2[j].length <= 30
  words1[i] and words2[j] All contain only lower case letters .
  source ： Power button （LeetCode）

Their thinking

   Count two... Respectively words The frequency of the string in , Then traverse the common parts of the two frequency tables , Search frequency is 1 String .

class Solution:
    def countWords(self, words1: List[str], words2: List[str]) -> int:
        words1=Counter(words1)
        words2=Counter(words2)
        count=0
        for i in words1.keys()&words2.keys():
            if words1[i]==1 and words2[i]==1:
                count+=1
        return count