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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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