[LeetCode] 893. Groups of Special-Equivalent Strings

 Problem : https://leetcode.com/problems/groups-of-special-equivalent-strings/

Because one character on odd position can be unlimitedly swapped with character on another odd position, the order of characters on odd position does not matter. Same to characters on even position. Two words are special-equivalent when they have same set of characters on odd positions and same set of characters on even positions.

class Solution {
    public int numSpecialEquivGroups(String[] words) {
        Set<String> group = new HashSet<>();
        
        for (String w: words) {
            group.add(keyOf(w));
        }
        
        return group.size();
    }
    
    String keyOf(String word) {
        int[] countOdd = new int[26];
        int[] countEven = new int[26];
        
        for (int i = 0; i < word.length(); i++) {
            if ((i & 1) == 1) {
                // odd position
                countOdd[word.charAt(i) - 'a'] += 1;
            } else {
                // even position
                countEven[word.charAt(i) - 'a'] += 1;
            }
        }
        
        StringBuilder sb = new StringBuilder();
        sb.append(Arrays.toString(countOdd));
        sb.append("-");
        sb.append(Arrays.toString(countEven));
       
        return sb.toString();
    }
}