How do you solve Groups of Special-Equivalent Strings on LeetCode?

Groups of Special-Equivalent Strings (LeetCode #893) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The even and odd indexed characters can be treated separately, allowing us to simplify the problem.

What is the brute force solution for Groups of Special-Equivalent Strings?

The brute force approach for Groups of Special-Equivalent Strings is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach checks every pair of strings to see if they can be transformed into each other by swapping characters at even and odd indices. This is straightforward but inefficient as it requires checking all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Groups of Special-Equivalent Strings?

Groups of Special-Equivalent Strings uses the following concepts: Array, Hash Table, String, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Groups of Special-Equivalent Strings?

Clearly explain your thought process and how you plan to group the strings. Consider edge cases, such as strings of different lengths or identical strings. Practice explaining your code as you write it, as communication is key in interviews.

What are common mistakes when solving Groups of Special-Equivalent Strings?

Failing to consider the order of characters when grouping, leading to incorrect counts. Not using a set to ensure uniqueness, which can result in over-counting groups.

#893

Groups of Special-Equivalent Strings

Medium

ArrayHash TableStringSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution leverages the idea of grouping strings by their sorted even and odd indexed characters. By creating a unique representation for each string based on its even and odd characters, we can efficiently count the number of special-equivalent groups.

⚙️

Algorithm

4 steps

1Step 1: Initialize a set to store unique representations of the strings.
2Step 2: For each string, extract and sort the characters at even indices and odd indices.
3Step 3: Combine the sorted even and odd characters into a tuple and add it to the set.
4Step 4: The size of the set at the end will give the number of special-equivalent groups.

solution.py10 lines

1# Full working Python code
2from collections import defaultdict
3
4def numSpecialEquivGroups(words):
5    unique_forms = set()
6    for word in words:
7        even_chars = ''.join(sorted(word[i] for i in range(0, len(word), 2)))
8        odd_chars = ''.join(sorted(word[i] for i in range(1, len(word), 2)))
9        unique_forms.add((even_chars, odd_chars))
10    return len(unique_forms)

ℹ

Complexity note: The time complexity is O(n) because we process each string in linear time to extract and sort even and odd characters. The space complexity is O(n) for storing unique forms in the set.

1The even and odd indexed characters can be treated separately, allowing us to simplify the problem.
2Sorting the characters at even and odd indices creates a unique representation for each string.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.