Medium
You are given an array of strings of the same length words
.
In one move, you can swap any two even indexed characters or any two odd indexed characters of a string words[i]
.
Two strings words[i]
and words[j]
are special-equivalent if after any number of moves, words[i] == words[j]
.
words[i] = "zzxy"
and words[j] = "xyzz"
are special-equivalent because we may make the moves "zzxy" -> "xzzy" -> "xyzz"
.A group of special-equivalent strings from words
is a non-empty subset of words such that:
words[i]
not in the group such that words[i]
is special-equivalent to every string in the group).Return the number of groups of special-equivalent strings from words
.
Example 1:
Input: words = [“abcd”,”cdab”,”cbad”,”xyzz”,”zzxy”,”zzyx”]
Output: 3
Explanation:
One group is [“abcd”, “cdab”, “cbad”], since they are all pairwise special equivalent, and none of the other
strings is all pairwise special equivalent to these.
The other two groups are [“xyzz”, “zzxy”] and [“zzyx”].
Note that in particular, “zzxy” is not special equivalent to “zzyx”.
Example 2:
Input: words = [“abc”,”acb”,”bac”,”bca”,”cab”,”cba”]
Output: 3
Constraints:
1 <= words.length <= 1000
1 <= words[i].length <= 20
words[i]
consist of lowercase English letters.import java.util.HashSet;
public class Solution {
public int numSpecialEquivGroups(String[] words) {
HashSet<String> set = new HashSet<>();
int result = 0;
for (String str : words) {
if (set.add(getHashBySwap(str.toCharArray()))) {
result++;
}
}
return result;
}
private String getHashBySwap(char[] chars) {
for (int i = 0; i < chars.length; i++) {
int j = i + 2;
while (j < chars.length) {
if (chars[i] > chars[j]) {
char temp = chars[j];
chars[j] = chars[i];
chars[i] = temp;
}
j += 2;
}
}
return String.valueOf(chars);
}
}