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Groups of Special-Equivalent Strings

Number: 929

Difficulty: Medium

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Companies: Meta

Problem Description

Given an array of strings of equal length, determine the number of groups of special-equivalent strings. Two strings are considered special-equivalent if by repeatedly swapping any two characters located at even indices or any two characters at odd indices, one string can be transformed into the other.

Key Insights

  • Swaps on even and odd positions are independent.
  • The final order of characters on even positions can be arranged arbitrarily (and similarly for odd positions).
  • Thus, the relative frequency (or sorted order) of characters at even indices and odd indices defines an invariant signature.
  • Two strings are special-equivalent if and only if their even-index characters (sorted) and odd-index characters (sorted) are identical.

Space and Time Complexity

Time Complexity: O(n * k log k), where n is the number of words and k is the length of each word (due to sorting). Space Complexity: O(n * k), for storing the signature of each word.

Solution

The solution iterates through each word and partitions its characters into even and odd index groups. Each group is sorted to generate a canonical signature (e.g., sorted even characters concatenated with sorted odd characters). By storing these signatures in a set, duplicate signatures (which represent special-equivalent words) are automatically filtered out. Finally, the number of unique signatures in the set corresponds to the number of groups.

Code Solutions

# Python solution with detailed comments

class Solution:
    def numSpecialEquivGroups(self, words: List[str]) -> int:
        groups = set()  # Set to store unique signatures for each group
        for word in words:
            even_chars = []  # Characters at even indices
            odd_chars = []   # Characters at odd indices
            for i, char in enumerate(word):
                if i % 2 == 0:
                    even_chars.append(char)  # Add to even characters list
                else:
                    odd_chars.append(char)   # Add to odd characters list
            # Sort the characters to create a canonical representation
            even_chars.sort()
            odd_chars.sort()
            signature = "".join(even_chars) + "".join(odd_chars)
            groups.add(signature)  # Add the signature to the set
        return len(groups)  # Number of unique signatures equals the number of groups
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