893. Groups of Special-Equivalent Strings

#### QUESTION:

You are given an array A of strings.

Two strings S and T are special-equivalent if after any number of moves, S == T.

A move consists of choosing two indices i and j with i % 2 == j % 2, and swapping S[i] with S[j].

Now, a group of special-equivalent strings from A is a non-empty subset S of A 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 from A.

Example 1:

Input: ["a","b","c","a","c","c"]
Output: 3
Explanation: 3 groups ["a","a"], ["b"], ["c","c","c"]

Example 2:

Input: ["aa","bb","ab","ba"]
Output: 4
Explanation: 4 groups ["aa"], ["bb"], ["ab"], ["ba"]

Example 3:

Input: ["abc","acb","bac","bca","cab","cba"]
Output: 3
Explanation: 3 groups ["abc","cba"], ["acb","bca"], ["bac","cab"]

Example 4:

Output: 1

Note:

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

#### SOLUTION:

class Solution {
public int numSpecialEquivGroups(String[] A) {
Set<Map<Integer,Integer>> result = new HashSet<>();
for(int i = 0;i<A.length;i++){
char[] chars = A[i].toCharArray();
HashMap<Integer,Integer> map = new HashMap<>();
for(int j = 0;j<chars.length;j+=2){
map.put(chars[j]+0,map.get(chars[j]+0)==null?1:map.get(chars[j]+0)+1);
}
for(int j = 1;j<chars.length;j+=2){
map.put(chars[j]-32,map.get(chars[j]-32)==null?1:map.get(chars[j]-32)+1);
}