Leetcode simple question: check whether two strings are almost equal

subject

If two strings word1 and word2 In the from ‘a’ To ‘z’ The difference in the frequency of each letter is No more than 3 , So we call these two strings word1 and word2 Almost equal .
Here you are. Both lengths are n String word1 and word2 , If word1 and word2 Almost equal , Please return true , Otherwise return to false .
A letter x Appearance frequency It refers to the number of times it appears in the string .
Example 1：
Input ：word1 = “aaaa”, word2 = “bccb”
Output ：false
explain ： character string “aaaa” There is 4 individual ‘a’ , however “bccb” There is 0 individual ‘a’ .
The difference between the two is 4 , Greater than upper limit 3 .
Example 2：
Input ：word1 = “abcdeef”, word2 = “abaaacc”
Output ：true
explain ：word1 and word2 The difference in the frequency of each letter in the is at most 3 ：

• ‘a’ stay word1 In the 1 Time , stay word2 In the 4 Time , The difference is 3 .
• ‘b’ stay word1 In the 1 Time , stay word2 In the 1 Time , The difference is 0 .
• ‘c’ stay word1 In the 1 Time , stay word2 In the 2 Time , The difference is 1 .
• ‘d’ stay word1 In the 1 Time , stay word2 In the 0 Time , The difference is 1 .
• ‘e’ stay word1 In the 2 Time , stay word2 In the 0 Time , The difference is 2 .
• ‘f’ stay word1 In the 1 Time , stay word2 In the 0 Time , The difference is 1 .
  Example 3：
  Input ：word1 = “cccddabba”, word2 = “babababab”
  Output ：true
  explain ：word1 and word2 The difference in the frequency of each letter in the is at most 3 ：
• ‘a’ stay word1 In the 2 Time , stay word2 In the 4 Time , The difference is 2 .
• ‘b’ stay word1 In the 2 Time , stay word2 In the 5 Time , The difference is 3 .
• ‘c’ stay word1 In the 3 Time , stay word2 In the 0 Time , The difference is 3 .
• ‘d’ stay word1 In the 2 Time , stay word2 In the 0 Time , The difference is 2 .
  Tips ：
  n == word1.length == word2.length
  1 <= n <= 100
  word1 and word2 All contain only lower case letters .
  source ： Power button （LeetCode）

Their thinking

   The almost equal strings given in the title are based on whether the frequency of characters appears differently , So the first step is to count the frequency of each character in each string , Then compare them one by one .

class Solution:
    def checkAlmostEquivalent(self, word1: str, word2: str) -> bool:
        word1=Counter(word1)
        word2=Counter(word2)
        for i in word1.keys()|word2.keys(): # In string 1 Or string 2 All the letters that have appeared 
            if abs(word1[i]-word2[i])>3:
                return False
        return True