Solution Explanation

This problem asks us to find the number of groups of special-equivalent strings. Two strings are special-equivalent if one can be transformed into the other by swapping even-indexed characters or odd-indexed characters independently. The core idea is to create a unique signature for each equivalence class.

Algorithm:

1. Separate Even and Odd Indices: For each string, we separate the characters at even and odd indices into two separate strings.

2. Sort: We sort the characters within each of these substrings (even and odd). Sorting ensures that the order of characters within the even or odd positions doesn't matter.

3. Concatenate and Hash: Concatenate the sorted even-indexed substring and the sorted odd-indexed substring. This creates a unique signature for each equivalence class. We use a Set (or unordered_set in C++) to store these signatures, ensuring that we only count each unique equivalence class once.

4. Return Size: The size of the Set (number of unique signatures) represents the number of groups of special-equivalent strings.

Time Complexity Analysis:

• Sorting: Sorting each substring takes O(k log k) time, where k is the length of the substring (approximately half the length of the word). Since we do this for each word, the total time complexity related to sorting is O(n * k log k), where n is the number of words.

• Other operations: Separating even and odd indices, concatenating, and set insertion all take linear time with respect to the length of the word and the number of words. Thus, their contribution is O(n*k).

• Overall: The dominant factor is the sorting step, making the overall time complexity O(n * k log k).

Space Complexity Analysis:

The space complexity is dominated by the Set (or unordered_set), which in the worst case stores all unique signatures of the words (n in the worst case). The space used for storing temporary strings and lists is also proportional to the size of the input. Therefore, the space complexity is O(n*k).

Code Explanation (Python):

class Solution:
    def numSpecialEquivGroups(self, words: List[str]) -> int:
        s = {''.join(sorted(word[::2]) + sorted(word[1::2])) for word in words}
        return len(s)
 

This Python code is highly concise. It leverages Python's list slicing (word[::2] for even indices, word[1::2] for odd indices) and set comprehension for efficient implementation. sorted() sorts the substrings in place, and ''.join(...) concatenates them. The set automatically handles uniqueness.

Code Explanation (Java):

The Java code is more verbose but follows the same logic:

class Solution {
    public int numSpecialEquivGroups(String[] words) {
        Set<String> s = new HashSet<>();
        for (String word : words) {
            s.add(convert(word));
        }
        return s.size();
    }
 
    private String convert(String word) {
        // ... (Implementation of separating, sorting, and concatenating as described above) ...
    }
}

The convert method handles the detailed steps of separating, sorting, and concatenating for each word.

The C++, Go, and other language solutions would follow a similar structure, adapting the syntax to the specifics of their respective languages, but maintaining the same fundamental algorithm.