You are given an array A of strings.
Two strings S and T are special-equivalent if after any number of moves, S == T.
A move consists of choosing two indices i and j with i % 2 == j % 2, and swapping S[i] with S[j].
Now, a group of special-equivalent strings from A is a non-empty subset S of A such that any string not in S is not special-equivalent with any string in S.
Return the number of groups of special-equivalent strings from A.
Example 1:
Input: ["a","b","c","a","c","c"]
Output: 3
Explanation: 3 groups ["a","a"], ["b"], ["c","c","c"]
Example 2:
Input: ["aa","bb","ab","ba"]
Output: 4
Explanation: 4 groups ["aa"], ["bb"], ["ab"], ["ba"]
Example 3:
Input: ["abc","acb","bac","bca","cab","cba"]
Output: 3
Explanation: 3 groups ["abc","cba"], ["acb","bca"], ["bac","cab"]
Example 4:
Input: ["abcd","cdab","adcb","cbad"]
Output: 1
Explanation: 1 group ["abcd","cdab","adcb","cbad"]
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):
"""
:type A: List[str]
:rtype: int
"""
s =set()
for t in A:
even = ''
odd = ''
for i in range(len(t)):
if i%2==0:
even += t[i]
else:
odd += t[i]
s.add(''.join(sorted(even))+''.join(sorted(odd)))
return len(s)