Groups of SpecialEquivalent Strings
To find the number of groups of specialequivalent strings, we can represent each string by a specific pattern and then use a set to count the unique patterns.
A pattern for a specialequivalent string can be represented by sorting the characters at even indices and sorting the characters at odd indices. This is because we can only swap characters at even indices with each other and characters at odd indices with each other.
Here’s the code:
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 class Solution:
def numSpecialEquivGroups(self, words: List[str]) > int:
# Create a set to store unique patterns of specialequivalent strings
special_equiv_groups = set()
# Iterate through each word and find its pattern
for word in words:
# Separate even and odd indexed characters
even_chars = ''.join(word[i] for i in range(0, len(word), 2))
odd_chars = ''.join(word[i] for i in range(1, len(word), 2))
# Sort the even and odd indexed characters and concatenate to form the pattern
pattern = ''.join(sorted(even_chars)) + ''.join(sorted(odd_chars))
# Add the pattern to the set of specialequivalent groups
special_equiv_groups.add(pattern)
# The number of unique patterns represents the number of groups of specialequivalent strings
return len(special_equiv_groups)

This code works by defining a pattern for each specialequivalent string and counting the unique patterns. The time complexity is O(n * m * log(m)), where n is the number of words, and m is the length of each word. The space complexity is O(n * m), as we are storing a pattern for each word in the input list.