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.

• 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++]);
}
return set.size();
}
}

############

class Solution {
public int numSpecialEquivGroups(String[] words) {
Set<String> s = new HashSet<>();
for (String word : words) {
}
return s.size();
}

private String convert(String word) {
List<Character> a = new ArrayList<>();
List<Character> b = new ArrayList<>();
for (int i = 0; i < word.length(); ++i) {
char ch = word.charAt(i);
if (i % 2 == 0) {
} else {
}
}
Collections.sort(a);
Collections.sort(b);
StringBuilder sb = new StringBuilder();
for (char c : a) {
sb.append(c);
}
for (char c : b) {
sb.append(c);
}
return sb.toString();
}
}

• 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:
return len(B)

• func numSpecialEquivGroups(words []string) int {
s := map[string]bool{}
for _, word := range words {
a, b := []rune{}, []rune{}
for i, c := range word {
if i&1 == 1 {
a = append(a, c)
} else {
b = append(b, c)
}
}
sort.Slice(a, func(i, j int) bool {
return a[i] < a[j]
})
sort.Slice(b, func(i, j int) bool {
return b[i] < b[j]
})
s[string(a)+string(b)] = true
}
return len(s)
}

• class Solution {
public:
int numSpecialEquivGroups(vector<string>& words) {
unordered_set<string> s;
for (auto& word : words) {
string a = "", b = "";
for (int i = 0; i < word.size(); ++i) {
if (i & 1)
a += word[i];
else
b += word[i];
}
sort(a.begin(), a.end());
sort(b.begin(), b.end());
s.insert(a + b);
}
return s.size();
}
};