[LeetCode] 1657. Determine if Two Strings Are Close

1. Question

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"

2. Thoughts

3. Tips learned

4. My solution

class Solution:
    def closeStrings(self, word1: str, word2: str) -> bool:
        if set(word1) != set(word2):
            return False
        return Counter(Counter(word1).values()) == Counter(Counter(word2).values())

153/153 cases passed (127 ms)
Your runtime beats 80.34 % of python3 submissions
Your memory usage beats 98.93 % of python3 submissions (17.4 MB)

5. AI Solution and Improvements

Your code is actually quite Pythonic already. However, it does perform the counting operation twice which is not necessary. Here's a slightly optimized version of your solution:

from collections import Counter

class Solution:
    def closeStrings(self, word1: str, word2: str) -> bool:
        counter1, counter2 = Counter(word1), Counter(word2)
        return set(word1) == set(word2) and sorted(counter1.values()) == sorted(counter2.values())

In this optimized version:

• We store the counts of each character in word1 and word2 in counter1 and counter2 respectively, so we don't have to create a Counter object twice for each word.
• We then check if the sets of characters in both words are equal.
• Finally, we compare the sorted lists of counts instead of converting them into Counters again. If the sorted counts are equal, then the words have the same frequency of each character.

This solution is both efficient and Pythonic.