Problem Description
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]. 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. Return the number of groups of special-equivalent strings from words.
Key Insights
- Characters at even indices can be rearranged independently of characters at odd indices.
- To determine if two strings are special-equivalent, we can check if their even-indexed and odd-indexed characters can form the same multiset.
- We can represent each string by a tuple of sorted even-indexed characters and sorted odd-indexed characters.
- All strings that result in the same tuple belong to the same group.
Space and Time Complexity
Time Complexity: O(n * m log m), where n is the number of strings and m is the length of each string (since we sort the characters). Space Complexity: O(n), for storing the unique groups in a set.
Solution
To solve this problem, we will iterate through each string in the words array. For each string, we will separate the characters into even-indexed and odd-indexed characters, sort them, and create a tuple. This tuple will serve as a unique identifier for the group of special-equivalent strings. By using a set to store these unique tuples, we can easily count the number of distinct groups.