2068. Check whether the two strings are almost equal

2068. Check that the two strings are almost equal

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 .

This question is still very interesting , The solution code is as follows ：

bool checkAlmostEquivalent(char * word1, char * word2){
    
    int r[26];
    int i;
    for(i=0;i<26;i++){
    
        r[i]=0;
    }
    for(i=0;word1[i]!='\0';i++){
    
        r[word1[i]-'a']++;
    }
    for(i=0;word2[i]!='\0';i++){
    
        r[word2[i]-'a']--;
    }
  for(i=0;i<26;i++){
    
        if(abs(r[i])>3){
    
            return false;
        }
    }
    return true;
}