## 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.

Solution To The Above Question

So if the value of k=1 then we have to make all the possible strings and then choose which one is the smallest string and then return it.

For K>1, we can simply start making a sorted string from the end by choosing the smallest letter from the first K letter.

class Solution {

public:

string orderlyQueue(string s, int k) {

if(k==1){

string ans=s;

string temp=s;

for(int i=0; i<s.length(); i++){

temp=temp+temp[0];

temp.erase(temp.begin());

if(temp<ans){

ans=temp;

}

}

return ans;

}

sort(s.begin(),s.end());

return s;

}

};

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