[LeetCode] Check Whether Two Strings are Almost Equivalent

1.문제

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.

같은 길이의 문자열 word1, word2 가 주어질 때 word1 , word2의 각 알파벳의 개수의 차이가 3이내이면 true , 아니라면 false를 리턴하는 문제이다.

Example 1

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

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

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.

2.풀이

1. 두 문자열을 순회하면서 객체에 각 문자의 개수를 저장한다.
2. word1의 문자는 1씩 증가 , word2의 문자는 1씩 감소하게 한다.
3. 객체에서 각 value의 절대값이 3보다 작은지 체크한다.

/**
 * @param {string} word1
 * @param {string} word2
 * @return {boolean}
 */
const checkAlmostEquivalent = function (word1, word2) {
  let map = {};
  for (let i = 0; i < word1.length; i++) {
    // 두 배열을 순회하면서 word1의 문자의 개수는 1씩 증가 , word2의 문자는 1씩 감소 해서 객체에 저장한다
    if (!map[word1[i]]) {
      map[word1[i]] = 0;
    }

    if (!map[word2[i]]) {
      map[word2[i]] = 0;
    }

    map[word1[i]]++;
    map[word2[i]]--;
  }

  // 객체에서 value 값의 절댓값이 3보다 큰 경우가 있다면 false를 리턴한다
  return !Object.keys(map).find((letter) => Math.abs(map[letter]) > 3);
};

3.결과