leetcode - 1657. Determine if Two Strings Are Close

Description

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.

Solution

Don’t need to perform those 2 operations. Op1 means we don’t need to care about the position of each character, op2 means we don’t need to care which characters do we have. All we need to confirm is: if the occurrence of each element is the same.

Time complexity: $o(n)$
Space complexity: $o(n)$

Code

class Solution:
    def closeStrings(self, word1: str, word2: str) -> bool:
        word_cnt1, word_cnt2 = Counter(word1), Counter(word2)
        ocurr1, ocurr2 = Counter(word_cnt1.values()), Counter(word_cnt2.values())
        return ocurr1 == ocurr2 and set(word1) == set(word2)