893. Groups of Special-Equivalent Strings

Explanation

To solve this problem, we can iterate through each word in the input array and normalize each word by sorting the characters at even indices and characters at odd indices separately. We can then store these normalized words in a Set to keep track of unique special-equivalent groups. The size of this set will give us the number of groups of special-equivalent strings.

Algorithm:

Initialize a HashSet to store unique special-equivalent groups.
Iterate through each word in the input array.
For each word, split the characters into two parts: one for even indices and one for odd indices.
Sort the characters at even indices and odd indices separately.
Merge the sorted even indices and sorted odd indices to form the normalized word.
Add the normalized word to the HashSet.
Return the size of the HashSet as the number of groups of special-equivalent strings.

Time Complexity: O(N * K * log K), where N is the number of words and K is the length of each word. Sorting each word takes O(K * log K) time.

Space Complexity: O(N * K) for storing the normalized words in the HashSet.

View Leetcode Problem

class Solution {
    public int numSpecialEquivGroups(String[] words) {
        Set<String> groups = new HashSet<>();
        
        for (String word : words) {
            char[] even = new char[(word.length() + 1) / 2];
            char[] odd = new char[word.length() / 2];
            
            for (int i = 0; i < word.length(); i++) {
                if (i % 2 == 0) {
                    even[i / 2] = word.charAt(i);
                } else {
                    odd[i / 2] = word.charAt(i);
                }
            }
            
            Arrays.sort(even);
            Arrays.sort(odd);
            
            String normalized = new String(even) + new String(odd);
            groups.add(normalized);
        }
        
        return groups.size();
    }
}

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Explanation

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