Introduction

In this blog post, we will explore how to solve the LeetCode problem "893" 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

You are given an array of strings of the same length words. In one move, you can swap any two even indexed characters or any two odd indexed characters of a string words[i]. Two strings words[i] and words[j] are special-equivalent if after any number of moves, words[i] == words[j]. For example, words[i] = "zzxy" and words[j] = "xyzz" are special-equivalent because we may make the moves "zzxy" -> "xzzy" -> "xyzz". A group of special-equivalent strings from words is a non-empty subset of words such that: Every pair of strings in the group are special equivalent, and The group is the largest size possible (i.e., there is not a string words[i] not in the group such that words[i] is special-equivalent to every string in the group). Return the number of groups of special-equivalent strings from words. Example 1: Input: words = ["abcd","cdab","cbad","xyzz","zzxy","zzyx"] Output: 3 Explanation: One group is ["abcd", "cdab", "cbad"], since they are all pairwise special equivalent, and none of the other strings is all pairwise special equivalent to these. The other two groups are ["xyzz", "zzxy"] and ["zzyx"]. Note that in particular, "zzxy" is not special equivalent to "zzyx". Example 2: Input: words = ["abc","acb","bac","bca","cab","cba"] Output: 3 Constraints: 1 <= words.length <= 1000 1 <= words[i].length <= 20 words[i] consist of lowercase English letters. All the strings are of the same length.

Explanation

Here's the breakdown of the solution:

• Key Idea: Two strings are special-equivalent if the sorted characters at even indices are the same and the sorted characters at odd indices are the same.
• Normalization: For each word, extract the even-indexed characters and odd-indexed characters, sort them independently, and then concatenate them into a normalized string. This normalized string serves as a unique identifier for special-equivalence.

• Counting Groups: Use a set to store the unique normalized strings. The size of the set represents the number of special-equivalent groups.

• Complexity: O(N L log(L)) time, O(N * L) space, where N is the number of words and L is the length of each word.

Code

    def numSpecialEquivGroups(words):
    """
    Counts the number of special-equivalent groups of strings.

    Args:
        words: A list of strings.

    Returns:
        The number of special-equivalent groups.
    """
    normalized_words = set()
    for word in words:
        even_chars = sorted(word[::2])
        odd_chars = sorted(word[1::2])
        normalized_word = "".join(even_chars) + "".join(odd_chars)
        normalized_words.add(normalized_word)
    return len(normalized_words)