Orderly Queue.java

H
1536248915
tags: Math, String
```

/*
A string S of lowercase letters is given. Then, we may make any number of moves.

In each move, we choose one of the first K letters (starting from the left), remove it, and place it at the end of the string.

Return the lexicographically smallest string we could have after any number of moves.

 

Example 1:

Input: S = "cba", K = 1
Output: "acb"
Explanation: 
In the first move, we move the 1st character ("c") to the end, obtaining the string "bac".
In the second move, we move the 1st character ("b") to the end, obtaining the final result "acb".
Example 2:

Input: S = "baaca", K = 3
Output: "aaabc"
Explanation: 
In the first move, we move the 1st character ("b") to the end, obtaining the string "aacab".
In the second move, we move the 3rd character ("c") to the end, obtaining the final result "aaabc".
 

Note:

1 <= K <= S.length <= 1000
S consists of lowercase letters only.
*/


/*
Thoughts:
Figure out: 
when k = 1, we can simply rotate and compare
when k > 1, we can just sort entire string
*/
class Solution {
 public String orderlyQueue(String S, int K) {
 if (K > 1) {
 char[] arr = S.toCharArray();
 Arrays.sort(arr);
 return new String(arr);
 }
 String res = S;
 for (int i = 1; i < S.length(); i++) { // move front section, one letter at a time
 String tmp = S.substring(i) + S.substring(0, i);
 if (res.compareTo(tmp) > 0) res = tmp;
 }
 return res;
 }
}
```