Introduction

In this blog post, we will explore how to solve the LeetCode problem "1657" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's a breakdown of the approach and the Python code:

• Key Idea: Two strings are "close" if they have the same set of characters and the same frequency distribution of those characters. The order of the characters themselves doesn't matter because of the swap operation. The character replacement operation allows us to essentially remap the characters based on their frequencies.
• Approach:
  • Check if the set of characters in both strings is the same. If not, they can't be close.
  • Count the frequency of each character in both strings.
  • Sort the frequency counts for both strings. If the sorted frequency counts are the same, the strings are close.
• Complexity: O(N) time, O(1) space. N is the length of the strings. The space complexity is O(1) because we are only using arrays of fixed size (26 for the English alphabet).

Code

    def closeStrings(word1: str, word2: str) -> bool:
    """
    Determines if two strings are close based on the defined operations.
    """

    if len(word1) != len(word2):
        return False

    freq1 = {}
    freq2 = {}

    for char in word1:
        freq1[char] = freq1.get(char, 0) + 1

    for char in word2:
        freq2[char] = freq2.get(char, 0) + 1

    if set(freq1.keys()) != set(freq2.keys()):
        return False

    freq_values1 = sorted(freq1.values())
    freq_values2 = sorted(freq2.values())

    return freq_values1 == freq_values2