# 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 <= 105`
• `word1` and `word2` contain only lowercase English letters.
 ``````1 2 3 4 5 6 7 8 `````` ``````class Solution: def closeStrings(self, word1: str, word2: str) -> bool: if len(word1) != len(word2) or set(word1) != set(word2): return False return ( sorted([word1.count(c) for c in set(word1)]) == sorted([word2.count(c) for c in set(word2)]) )``````