［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();
}
}