899. Orderly Queue

899. Orderly Queue (Hard)

You are given a string s and an integer k. You can choose one of the first k letters of s and append it at the end of the string..

Return the lexicographically smallest string you could have after applying the mentioned step 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".

 

Constraints:

• 1 <= k <= s.length <= 1000
• s consist of lowercase English letters.

Companies:
Amazon

Related Topics:
Math, String, Sorting

Solution 1.

When K >= 2, we can in effect swap any two characters, which means we can do bubble sort. So the result for K >= 2 is just the sorted S.

When K == 1, we need to rotate the string and find the minimum.

// OJ: https://leetcode.com/problems/orderly-queue/
// Author: github.com/lzl124631x
// Time: O(S^2)
// Space: O(S)
class Solution {
public:
    string orderlyQueue(string S, int K) {
        if (K > 1) {
            sort(S.begin(), S.end());
            return S;
        }
        string ans = S;
        for (int i = 1; i < S.size(); ++i) {
            rotate(S.begin(), S.begin() + 1, S.end());
            if (S < ans) ans = S;
        }
        return ans;
    }
};