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
- For example,
- 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
(alla
's turn intob
's, and allb
's turn intoa
's)
- For example,
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.
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"
Input: word1 = "a", word2 = "aa" Output: false Explanation: It is impossible to attain word2 from word1, or vice versa, in any number of operations.
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"
1 <= word1.length, word2.length <= 105
word1
andword2
contain only lowercase English letters.
class Solution:
def closeStrings(self, word1: str, word2: str) -> bool:
if len(word1) != len(word2) or set(word1) != set(word2):
return False
count1 = [0] * 26
count2 = [0] * 26
for ch1, ch2 in zip(word1, word2):
count1[ord(ch1) - 97] += 1
count2[ord(ch2) - 97] += 1
return sorted(count1) == sorted(count2)