groups-of-special-equivalent-strings.py

# Time:  O(n * l)
# Space: O(n)

# 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(object):
    def numSpecialEquivGroups(self, A):
        """
        :type A: List[str]
        :rtype: int
        """
        def count(word):
            result = [0]*52
            for i, letter in enumerate(word):
                result[ord(letter)-ord('a') + 26*(i%2)] += 1
            return tuple(result)

        return len({count(word) for word in A})