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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0801_0900.s0899_orderly_queue;

// #Hard #String #Math #Sorting #2022_03_28_Time_1_ms_(100.00%)_Space_40.5_MB_(98.64%)

import java.util.ArrayList;
import java.util.Arrays;

/**
 * 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 1^st character 'c' to the end, obtaining the string "bac".
 *
 * In the second move, we move the 1^st 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 1^st character 'b' to the end, obtaining the string "aacab".
 *
 * In the second move, we move the 3^rd character 'c' to the end, obtaining the final result "aaabc". 
 *
 * **Constraints:**
 *
 * *   `1 <= k <= s.length <= 1000`
 * *   `s` consist of lowercase English letters.
**/
public class Solution {
    public String orderlyQueue(String s, int k) {
        if (k > 1) {
            char[] ans = s.toCharArray();
            Arrays.sort(ans);
            return String.valueOf(ans);
        }
        char min = 'z';
        ArrayList list = new ArrayList<>();
        for (int i = 0; i < s.length(); i++) {
            char cc = s.charAt(i);
            if (cc < min) {
                min = cc;
            }
        }
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == min) {
                list.add(i);
            }
        }
        String ans = s;
        for (Integer integer : list) {
            String after = s.substring(0, integer);
            String before = s.substring(integer);
            String f = before + after;
            if (f.compareTo(ans) < 0) {
                ans = f;
            }
        }
        return ans;
    }
}