"""
You are given an array A of strings.

A move onto S consists of swapping any two even indexed characters of S, or any two odd indexed characters of S.

Two strings S and T are special-equivalent if after any number of moves onto S, S == T.

For example, S = "zzxy" and T = "xyzz" are special-equivalent because we may make the moves "zzxy" -> "xzzy" -> "xyzz" that swap S[0] and S[2], then S[1] and S[3].

Now, a group of special-equivalent strings from A is a non-empty subset of A such that:

Every pair of strings in the group are special equivalent, and;
The group is the largest size possible (ie., there isn't a string S not in the group such that S is special equivalent to every string in the group)
Return the number of groups of special-equivalent strings from A.


Example 1:

Input: ["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 are 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: ["abc","acb","bac","bca","cab","cba"]
Output: 3

one line solution
from collections import Counter

class Solution:
  def numSpecialEquivGroups(self, A: List[str]) -> int:
    return len({tuple(tuple(sorted(Counter(S[j::2]).items())) for j in range(2)) for S in A})
"""

from collections import Counter


class Solution:
    def numSpecialEquivGroups(self, A) -> int:
        s = set()
        for S in A:
            evens = tuple(sorted(Counter(S[0::2]).items()))
            odds = tuple(sorted(Counter(S[1::2]).items()))
            s.add((evens, odds))
        return len(s)


if __name__ == '__main__':
    s = Solution()
    a = ["abcd", "cdab", "cbad", "xyzz", "zzxy", "zzyx"]
    res = s.numSpecialEquivGroups(a)
    print(res)
