893. Groups of Special-Equivalent Strings 奇数偶数位上的相同数

［抄题］：

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"]

[暴力解法]：

时间分析：

空间分析：

[优化后]：

时间分析：

空间分析：

［奇葩输出条件］：

［奇葩corner case］：

［思维问题］：

［英文数据结构或算法，为什么不用别的数据结构或算法］：

［一句话思路］：

［输入量］：空： 正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：

［画图］：

［一刷］：

［二刷］：

［三刷］：

［四刷］：

［五刷］：

[五分钟肉眼debug的结果]：

［总结］：

多开几个数组就行了

［复杂度］：Time complexity: O(n) Space complexity: O(n)

［算法思想：迭代/递归/分治/贪心］：

［关键模板化代码］：

［其他解法］：

［Follow Up］：

［LC给出的题目变变变］：

[代码风格] ：

[是否头一次写此类driver funcion的代码] ：

[潜台词] ：

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

      //initialization: 2 arrays and a set
      Set<String> set = new HashSet<>();
      for (String s : A) {
        int[] old = new int[26];
        int[] even = new int[26];
        for (int i = 0; i < s.length(); i++) {
          //add to the old or even array
          if (i % 2 == 0) {
            even[s.charAt(i) - 'a']++;
          } else {
            old[s.charAt(i) - 'a']++;
          }
        }
        String sig = Arrays.toString(old) + Arrays.toString(even);
        set.add(sig);
      }
/*
    acb
old [0, 0, 1...]
even [1, 1, 0...]

  bca
old[0, 0, 1...]
even[1, 1, 0...]
*/

      //return
      return set.size();
    }
}