[LeetCode] 893. Groups of Special-Equivalent Strings

Problem

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:

Input: [“abcd”,”cdab”,”adcb”,”cbad”]
Output: 1
Explanation: 1 group [“abcd”,”cdab”,”adcb”,”cbad”]

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

Thought: we only need to find the number of groups special permutations of the given string:
if two strings have all odd digits equivalent, and all even digits equivalent, they are in the same group.

class Solution {
    public int numSpecialEquivGroups(String[] A) {
        Set<String> set = new HashSet<>();
        for (String s: A) {
            int[] odd = new int[26];
            int[] even = new int[26];
            for (int i = 0; i < s.length(); i++) {
                int index = s.charAt(i)-`a`;
                if (i%2 == 1) odd[index]++;
                else even[index]++;
            }
            String record = Arrays.toString(odd) + Arrays.toString(even);
            set.add(record);
        }
        return set.size();
    }
}