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:
- Every pair of strings in the group are special equivalent, and;
- 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(); } };
No comments:
Post a Comment