893. Groups of Special-Equivalent Strings ----------E

Q:

You are given an array A of strings.

A move onto S consists of swapping any two even indexed characters of S, or any two odd indexed characters of S.

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

For example, S = "zzxy" and T = "xyzz" are special-equivalent because we may make the moves "zzxy" -> "xzzy" -> "xyzz" that swap S[0] and S[2], then S[1] and S[3].

Now, a group of special-equivalent strings from A is a non-empty subset of A such that:

1. Every pair of strings in the group are special equivalent, and;
2. The group is the largest size possible (ie., there isn't a string S not in the group such that S is special equivalent to every string in the group)

Return the number of groups of special-equivalent strings from A.

 

Example 1:

Input: ["abcd","cdab","cbad","xyzz","zzxy","zzyx"]
Output: 3
Explanation: 
One group is ["abcd", "cdab", "cbad"], since they are all pairwise special equivalent, and none of the other strings are all pairwise special equivalent to these.

The other two groups are ["xyzz", "zzxy"] and ["zzyx"].  Note that in particular, "zzxy" is not special equivalent to "zzyx".

Example 2:

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

 

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.

A:

class Solution {
public:
    int numSpecialEquivGroups(vector<string>& A) {
        unordered_set<string> S;
        for(auto  str : A){
            int n = str.length();
            vector<char> odd;
            vector<char> even;
            for(int i =0;i<n;i++){
                if(i%2==0){
                    odd.push_back(str[i]);
                }else{
                    even.push_back(str[i]);
                }
            }
            sort(odd.begin(), odd.end());
            sort(even.begin(),even.end());
            odd.insert(odd.end(), even.begin(), even.end());
            string key(odd.begin(), odd.end());
            S.insert(key);
        }
        return S.size();
    }
};