Is Determine if Two Strings Are Close easy or hard?

Determine if Two Strings Are Close is classified as a Medium difficulty problem. The challenge lies in recognizing that the operations allow arbitrary swaps and frequency exchanges, which reduces the problem to comparing character sets and frequency distributions.

Determine if Two Strings Are Close Python/Java solution

In Python, developers typically use collections.Counter or arrays to count characters and compare sorted frequency values. In Java, HashMap or int[26] arrays are common for counting characters and verifying frequency distributions efficiently.

How to solve Determine if Two Strings Are Close in O(n)?

Count the frequency of each character in both strings using arrays or hash maps. Verify that both strings contain the same set of characters. Then compare their frequency distributions using a hash-based comparison instead of sorting. This keeps the algorithm O(n) with O(k) extra space.

What is the best approach for Determine if Two Strings Are Close?

The most common approach compares the set of characters and the distribution of their frequencies. First check that both strings contain the same unique characters. Then compare the sorted frequency counts. If both conditions match, the strings are close. This runs in O(n + k log k) time where k is the alphabet size.

Is Determine if Two Strings Are Close asked at Google/Amazon/Meta?

String frequency and transformation problems like this appear frequently in interviews at companies such as Amazon, Meta, and Google. They test understanding of hash maps, counting techniques, and reasoning about allowed operations on strings.

What data structure is used in Determine if Two Strings Are Close?

The solution primarily uses hash tables or fixed-size frequency arrays to count characters. Sorting may also be used to compare frequency distributions. These structures allow efficient counting and constant-time lookups.

What is the time complexity of Determine if Two Strings Are Close?

The optimal solution runs in O(n) time for counting characters plus a small overhead based on the alphabet size. Using sorted frequency comparison gives O(n + k log k) time. Since k is at most 26 for lowercase letters, both approaches are effectively linear.

Description Hints 2 Solution Video 24:23

1657. Determine if Two Strings Are Close

Medium54.2% acceptance7 companies

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"