Problem URL: Groups of special equivalent strings

We will take each word, separate the even and odd characters, and sort them. Then we add them to a set. The size of the set is the number of groups.

class Solution:
    def numSpecialEquivGroups(self, words: List[str]) -> int:
        res = set()
        for w in words:
            sorted_odd_even = ''.join(sorted(w[1::2]) + sorted(w[::2]))
            res.add(sorted_odd_even)
        return len(res)

Time Complexity: O(n*klog(k)), n is the number of words, k is the lenght of longest word
Space Complexity: O(n*k)