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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
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.

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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++]);
                set.add(sorted.toString());
            }
            return set.size();
        }
    }
    
  • Todo
    
  • class Solution:
        def numSpecialEquivGroups(self, words: List[str]) -> int:
            s = {''.join(sorted(word[::2]) + sorted(word[1::2])) for word in words}
            return len(s)
    
    ############
    
    class Solution(object):
        def numSpecialEquivGroups(self, A):
            """
            :type A: List[str]
            :rtype: int
            """
            B = set()
            for a in A:
                B.add(''.join(sorted(a[0::2])) + ''.join(sorted(a[1::2])))
            return len(B)
    

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