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.

Solution

class Solution {
public:
    int numSpecialEquivGroups(vector<string>& A) { 
      std::unordered_set<std::string> myset;
      for (string x : A) {
          std::string odd, even;
          for(int i = 0; i < x.size(); ++i) {
              if(i%2 == 0) even += x[i];
              else odd += x[i];
          }
          sort(odd.begin(),odd.end());
          sort(even.begin(), even.end());
          myset.insert(odd+even);
      }
      return myset.size();
    }
};
By guozetang            Updated: 2020-09-19 13:02:30

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