899. Orderly Queue
Explanation:
The key observation for this problem is that:
- If k is greater than 1, we can rearrange the letters in any order we want.
- If k is equal to 1, we can only rotate the string to find the lexicographically smallest string.
Therefore, when k is greater than 1, the solution is simply to sort the string. When k is equal to 1, we need to try all rotations of the string to find the lexicographically smallest string.
class Solution {
public String orderlyQueue(String s, int k) {
if (k > 1) {
char[] arr = s.toCharArray();
Arrays.sort(arr);
return new String(arr);
} else {
String smallest = s;
for (int i = 1; i < s.length(); i++) {
String rotated = s.substring(i) + s.substring(0, i);
if (rotated.compareTo(smallest) < 0) {
smallest = rotated;
}
}
return smallest;
}
}
}Code Editor (Testing phase)
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