Algorithm
Problem Name: 899. Orderly Queue
Problem Link: https://leetcode.com/problems/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 <= 1000sconsist of lowercase English letters.
Code Examples
#1 Code Example with Javascript Programming
Code -
Javascript Programming
const orderlyQueue = function (S, K) {
if (K === 0) return S
else if (K > 1) return S.split('').sort().join('')
let result = 0,
L = S.length
for (let i = 1; i < L; i++) {
for (let j = 0; j < L; j++) {
let d = S.charCodeAt((result + j) % L) - S.charCodeAt((i + j) % L)
if (d !== 0) {
if (d > 0) result = i
break
}
}
}
return S.slice(result) + S.slice(0, result)
}
Input
s = "cba", k = 1
Output
"acb"
#2 Code Example with Python Programming
Code -
Python Programming
class Solution:
def orderlyQueue(self, S, K):
return "".join(sorted(S)) if K > 1 else min(S[i:] + S[:i] for i in range(len(S)))
Input
s = "cba", k = 1
Output
"acb"