Formatted question description: https://leetcode.ca/all/893.html

# 893. Groups of Special-Equivalent Strings (Easy)

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


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.

Companies:

Related Topics:
String

## Solution 1.

// OJ: https://leetcode.com/problems/groups-of-special-equivalent-strings/

// Time: O(NWlogW) where N is length of A, W is max word length
// Space: O(A) where A is the length of all the contents in array A
class Solution {
public:
int numSpecialEquivGroups(vector<string>& A) {
unordered_set<string> s;
for (string str : A) {
string a, b;
for (int i = 0; i < str.size(); ++i) {
if (i % 2) a += str[i];
else b += str[i];
}
sort(a.begin(), a.end());
sort(b.begin(), b.end());
s.insert(a + b);
}
return s.size();
}
};


Java

class Solution {
public int numSpecialEquivGroups(String[] A) {
Set<String> set = new HashSet<String>();
for (String str : A) {
int length = str.length();
StringBuffer even = new StringBuffer();
StringBuffer odd = new StringBuffer();
for (int i = 0; i < length; i++) {
char c = str.charAt(i);
if (i % 2 == 0)
even.append(c);
else
odd.append(c);
}
char[] evenArray = even.toString().toCharArray();
char[] oddArray = odd.toString().toCharArray();
Arrays.sort(evenArray);
Arrays.sort(oddArray);
StringBuffer sorted = new StringBuffer();
int evenLength = evenArray.length, oddLength = oddArray.length;
int evenIndex = 0, oddIndex = 0;
while (oddIndex < oddLength) {
sorted.append(evenArray[evenIndex++]);
sorted.append(oddArray[oddIndex++]);
}
if (evenIndex < evenLength)
sorted.append(evenArray[evenIndex++]);