[LeetCode] 1657. Determine if Two Strings Are Close

- Problem

1657. Determine if Two Strings Are Close
Two strings are considered close if you can attain one from the other using the following operations:

• Operation 1: Swap any two existing characters.
  - For example, abcde -> aecdb

• Operation 2: Transform every occurrence of one existing character into another existing character, and do the same with the other character.

  	- For example, `aacabb -> bbcbaa` (all `a`'s turn into `b`'s, and all `b`'s turn into `a`'s)

You can use the operations on either string as many times as necessary.

Given two strings, word1 and word2, return true if word1 and word2 are close, and false otherwise.

Example 1:

Input: word1 = "abc", word2 = "bca"
Output: true
Explanation: You can attain word2 from word1 in 2 operations.
Apply Operation 1: "abc" -> "acb"
Apply Operation 1: "acb" -> "bca"

Example 2:

Input: word1 = "a", word2 = "aa"
Output: false
Explanation: It is impossible to attain word2 from word1, or vice versa, in any number of operations.

Example 3:

Input: word1 = "cabbba", word2 = "abbccc"
Output: true
Explanation: You can attain word2 from word1 in 3 operations.
Apply Operation 1: "cabbba" -> "caabbb"
Apply Operation 2: "caabbb" -> "baaccc"
Apply Operation 2: "baaccc" -> "abbccc"

Constraints:

• 1 <= word1.length, word2.length <= 10^5
• word1 and word2 contain only lowercase English letters.

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내 풀이

class Solution:
    def closeStrings(self, word1: str, word2: str) -> bool:
        word1_counter, word2_counter = map(Counter, (word1, word2))

        return word1_counter.keys() == word2_counter.keys() and Counter(word1_counter.values()) == Counter(word2_counter.values())

close임을 판별하기 위해서는 다음과 같은 두 가지 조건이 필요하다.

• 각 문자열에 등장하는 문자가 동일한가?
  - operation1을 만족하기 위함이다.

• 만약 동일하다면, 키 값과 상관 없이 임의 원소의 빈도 수가 동일한가?
  - operation2를 만족하기 위함이다.

- 결과