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:

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

EXPLANATION:

题目比较难以理解，题目的意思就是，一组string里面每个string有相同的长度，那么替换其中的偶数位，或者奇数位，如果能够拼凑出相同的字符，那么他们就算一组。

理解了题目的意思之后就可以将问题简化成，计算每个string的奇数位和偶数位置每个字符的个数。

这里就需要map来进行组合，同时，为了区分奇偶，可以选择变成两个map，或者由于题目指定了，当前的字符串中只存在小写字母，那么我们就可以用大写字母来标记奇数位置。

最后将获得的结果进行保存，因为相同的组合相当于一个组，那么我们只能保存一个，这里就需要用到set的数据结构。

那么代码如下：

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);
            }
            result.add(map);
        }
        return result.size();
    }
}
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