899. Orderly Queue LeetCode Solution

In this guide, you will get 899. Orderly Queue LeetCode Solution with the best time and space complexity. The solution to Orderly Queue problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Orderly Queue solution in C++
Orderly Queue solution in Java
Orderly Queue solution in Python
Additional Resources

899. Orderly Queue LeetCode Solution image

Problem Statement of Orderly Queue

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.

Complexity Analysis

Time Complexity: O(n^2)
Space Complexity: O(n)

899. Orderly Queue LeetCode Solution in C++

class Solution {
 public:
  string orderlyQueue(string s, int k) {
    if (k > 1) {
      ranges::sort(s);
      return s;
    }

    string ans = s;

    for (int i = 1; i < s.length(); ++i)
      ans = min(ans, s.substr(i) + s.substr(0, i));

    return ans;
  }
};
/* code provided by PROGIEZ */

899. Orderly Queue LeetCode Solution in Java

class Solution {
  public String orderlyQueue(String s, int k) {
    if (k > 1) {
      char[] chars = s.toCharArray();
      Arrays.sort(chars);
      return String.valueOf(chars);
    }

    String ans = s;

    for (int i = 1; i < s.length(); ++i) {
      String t = s.substring(i) + s.substring(0, i);
      if (ans.compareTo(t) > 0)
        ans = t;
    }

    return ans;
  }
}
// code provided by PROGIEZ

899. Orderly Queue LeetCode Solution in Python

class Solution:
  def orderlyQueue(self, s: str, k: int) -> str:
    return (''.join(sorted(s)) if k > 1
            else min(s[i:] + s[:i] for i in range(len(s))))
 # code by PROGIEZ