You are given an arrayAof strings.

Two stringsSandTare special-equivalent if after any number ofmoves, S == T.

A _move _consists of choosing two indicesiandjwithi % 2 == j % 2, and swappingS[i]withS[j].

Now, agroup of special-equivalent strings fromA is a non-empty subset S ofA such that any string not in S is not special-equivalent with any string in S.

Return the number of groups of special-equivalent strings fromA.

Note:

• 1 <= A.length <= 1000
• 1 <= A[i].length <= 20
• AllA[i]have the same length.
• AllA[i]consist of only lowercase letters.

这个题就是按照规则统计奇数index上的所有字符出现次数和偶数index上所有字符出现的次数，将所有equivalent的字符串map到同一个字符串，再统计map之后的字符串个数

class Solution {
    public int numSpecialEquivGroups(String[] A) {
        if(A == null || A.length == 0)
            return 0;

        HashSet<String> set = new HashSet<String>();
        for(String s: A) {
            set.add(encode(s));
        }

        return set.size();
    }

    public String encode(String s) {
        StringBuilder res = new StringBuilder();

        char[] even = new char[26];
        char[] odd = new char[26];

        for(int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);

            if(i % 2 == 0) {
                even[c - 'a']++;
            } else{
                odd[c - 'a']++;
            }
        }

        for(int i = 0; i < 26; i++) {
            if(even[i] != 0)
                res.append((char) ('a' + i)).append(even[i]);
        }
        res.append('|');
        for(int i = 0; i < 26; i++) {
            if(odd[i] != 0)
                res.append((char) ('a' + i)).append(odd[i]);
        }

        return res.toString();
    }
}