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
Note:
1 <= A.length <= 1000
1 <= A[i].length <= 20
All A[i] have the same length.
All A[i] consist of only lowercase letters.
class Solution:
    def numSpecialEquivGroups(self, A: List[str]) -> int:
        return len(set(''.join(sorted(s[0::2])) + ''.join(sorted(s[1::2])) for s in A))
            

class Solution:
    def numSpecialEquivGroups(self, A: List[str]) -> int:
        return len(set(''.join(sorted(s[0::2])) + ''.join(sorted(s[1::2])) for s in A))
            
