Check Whether Two Strings are Almost Equivalent | LeetCode 2068 | C

2023-09-11
#LeetCode #C #Hash Table #String #Counting
LeetCode

Two strings word1 and word2 are considered almost equivalent if the differences between the frequencies of each letter from ‘a’ to ‘z’ between word1 and word2 is at most 3.

Given two strings word1 and word2, each of length n, return true if word1 and word2 are almost equivalent, or false otherwise.

The frequency of a letter x is the number of times it occurs in the string.

Example 1:

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Input: word1 = "aaaa", word2 = "bccb"
Output: false
Explanation: There are 4 'a's in "aaaa" but 0 'a's in "bccb".
The difference is 4, which is more than the allowed 3.

Example 2:

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Input: word1 = "abcdeef", word2 = "abaaacc"
Output: true
Explanation: The differences between the frequencies of each letter in word1 and word2 are at most 3:
- 'a' appears 1 time in word1 and 4 times in word2. The difference is 3.
- 'b' appears 1 time in word1 and 1 time in word2. The difference is 0.
- 'c' appears 1 time in word1 and 2 times in word2. The difference is 1.
- 'd' appears 1 time in word1 and 0 times in word2. The difference is 1.
- 'e' appears 2 times in word1 and 0 times in word2. The difference is 2.
- 'f' appears 1 time in word1 and 0 times in word2. The difference is 1.

Example 3:

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Input: word1 = "cccddabba", word2 = "babababab"
Output: true
Explanation: The differences between the frequencies of each letter in word1 and word2 are at most 3:
- 'a' appears 2 times in word1 and 4 times in word2. The difference is 2.
- 'b' appears 2 times in word1 and 5 times in word2. The difference is 3.
- 'c' appears 3 times in word1 and 0 times in word2. The difference is 3.
- 'd' appears 2 times in word1 and 0 times in word2. The difference is 2.

Constraints:

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n == word1.length == word2.length
1 <= n <= 100
word1 and word2 consist only of lowercase English letters.

Approach

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Calculate the frequency. If the diffrence between two words are higher than 4, return false.

Algorithm

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1. Go through the list and calculate the frequencies of two arrays.
2. Compare the two hash tables.

Implementation

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bool checkAlmostEquivalent(char * word1, char * word2){
    int table1[26] = {0}, table2[26] ={0};
    for(int i = 0;i<strlen(word1);i++){
        table1[word1[i]-'a']++;
        table2[word2[i]-'a']++;
    }
    for(int i = 0;i<26;i++){
        if(abs(table1[i]-table2[i])>3)
            return false;
    }
    return true;
}