893. Groups of Special-Equivalent Strings*
https://leetcode.com/problems/groups-of-special-equivalent-strings/
題目描述
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
not in the group such thatS
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
have the same length.A[i]
- All
consist of only lowercase letters.A[i]
C++ 實作 1
參考: C++ Simple Solution, 作者的想法是:
General Idea:
- Split strings in two to substrings, 1 with even indexed characters, and 1 with odd
- Sort the two substrings (We do this because if you can swap on string with another, when sorted they will equal each other because they must have the same characters)
- Insert your pair of strings into set, this will keep track of the unique “groups”
- Rerturn the size of your set
class Solution {
public:
int numSpecialEquivGroups(vector<string>& A) {
set<pair<string, string>> record;
for (auto &w : A) {
pair<string, string> p;
for (int i = 0; i < w.size(); ++ i) {
if (i % 2 == 0) p.first += w[i];
else p.second += w[i];
}
std::sort(p.first.begin(), p.first.end());
std::sort(p.second.begin(), p.second.end());
record.insert(p);
}
return record.size();
}
};