893

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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Related Topics:
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Solution 1.

// OJ: https://leetcode.com/problems/groups-of-special-equivalent-strings/
// Author: github.com/lzl124631x
// 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();
    }
};