Solved! Leetcode 2068. Check Whether Two Strings are Almost Equivalent

Description Check Whether Two Strings are Almost Equivalent

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

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

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

<strong>Input:</strong> word1 = "abcdeef", word2 = "abaaacc"
<strong>Output:</strong> true
<strong>Explanation:</strong> 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.

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

<strong>Input:</strong> word1 = "cccddabba", word2 = "babababab"
<strong>Output:</strong> true
<strong>Explanation:</strong> 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.

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

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

Solution

class Solution {
    public boolean checkAlmostEquivalent(String word1, String word2) {
        if(word1==null && word2==null){
            return true;
        }

        if(word1==null || word2==null){
            return false;
        }

        int[] f1 = getFreq(word1);
        int[] f2 = getFreq(word2);
        for(int i=0;i<f1.length;i++){
            if(Math.abs(f1[i]-f2[i])>3){
                return false;
            }
        }
        return true;
    }

    private int[] getFreq(String a){
        int[] freq = new int[26];
        for(int i=0;i<a.length();i++){
            freq[a.charAt(i)-'a']++;
        }
        return freq;
    }
}

class Solution {

public boolean checkAlmostEquivalent(String word1, String word2) {

if(word1==null && word2==null){

return true;

}

if(word1==null || word2==null){

return false;

}

int[] f1 = getFreq(word1);

int[] f2 = getFreq(word2);

for(int i=0;i<f1.length;i++){

if(Math.abs(f1[i]-f2[i])>3){

return false;

}

return true;

}

private int[] getFreq(String a){

int[] freq = new int[26];

for(int i=0;i<a.length();i++){

freq[a.charAt(i)-'a']++;

}

return freq;

}

Time Complexity

O(n), where n is the length of words

Space Complexity

O(26 * 2) -> O(1), constant space

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