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();
}
};