LeetCode in Kotlin

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.

Solution

class Solution {
    fun orderlyQueue(s: String, k: Int): String {
        if (k > 1) {
            val ans = s.toCharArray()
            ans.sort()
            return String(ans)
        }
        var min = 'z'
        val list = ArrayList<Int>()
        for (element in s) {
            if (element < min) {
                min = element
            }
        }
        for (i in s.indices) {
            if (s[i] == min) {
                list.add(i)
            }
        }
        var ans = s
        for (integer in list) {
            val after = s.substring(0, integer)
            val before = s.substring(integer)
            val f = before + after
            if (f < ans) {
                ans = f
            }
        }
        return ans
    }
}

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