超出圆形阵列旋转时间限制(TLE)

Question

I am really confused why my java code is not working it is giving TLE on Code Monks on Hacker Earth. Here is the link to the question - Link to Question the first question MONK AND ROTATION

import java.util.Scanner;
class TestClass {
    static int[] ar=new int[100001];
    public static void main(String args[] ){
        Scanner in=new Scanner(System.in);
        byte t=in.nextByte();
        while(--t>0){
            int n=in.nextInt();
            int k=in.nextInt()%n;
                for(int i=0;i<n-k;i++)
                    ar[i]=in.nextInt();
                for(int i=0;i<k;i++)
                    System.out.print(in.nextInt()+" ");
                for(int i=0;i<n-k;i++)
                    System.out.print(ar[i]+" ");
            System.out.println();
        }
    }
}

I don't know why is it giving TLE I think there is some infinite loop going.

the question at the site is-

Monk and Rotation Monk loves to perform different operations on arrays, and so being the principal of HackerEarth School, he assigned a task to his new student Mishki. Mishki will be provided with an integer array A of size N and an integer K , where she needs to rotate the array in the right direction by K steps and then print the resultant array. As she is new to the school, please help her to complete the task.

Input: The first line will consists of one integer T denoting the number of test cases. For each test case:

1. The first line consists of two integers N and K, N being the number of elements in the array and K denotes the number of steps of rotation.
2. The next line consists of N space separated integers , denoting the elements of the array A. Output: Print the required array.

Constraints:

1<=T<=20
1<=N<=10^5
0<=K<=10^6
0<=A[i]<=10^6

Sample Input

1

5 2

1 2 3 4 5

Sample Output

4 5 1 2 3

Explanation

Here T is 1, which means one test case.

denoting the number of elements in the array and , denoting the number of steps of rotations.

The initial array is: In first rotation, 5 will come in the first position and all other elements will move to one position ahead from their current position. Now, the resultant array will be

In second rotation, 4 will come in the first position and all other elements will move to one position ahead from their current position. Now, the resultant array will be

Time Limit: 1.0 sec(s) for each input file Memory Limit: 256 MB Source Limit: 1024 KB

Answer 1

Problem is here

for(int i=0;i<n-k;i++) //this loop you are using for storing input elements in array
    ar[i]=in.nextInt();

for(int i=0;i<k;i++) // here you again taking the input you don't need this loop
    System.out.print(in.nextInt()+" ");

for(int i=0;i<n-k;i++)
    System.out.print(ar[i]+" ");

You also need to change the condition in while loop while(--t>0) to while(--t >= 0) . It should be >= 0 other wise it will not work. Other solution is to use post decrement while(t-- > 0)

You are trying to print right rotation of the array. So you need to start printing the elements from index n - k . Then you need to calculate the end index it is (n - k) + n this is because array has n elements. Then you can access array elements like this arr[i % n] .

Here is the complete working solution

class TestClass {
    static int[] ar = new int[100001];

    public static void main(String args[]) {
        Scanner in = new Scanner(System.in);
        byte t = in.nextByte();
        while (--t >= 0) {
            int n = in.nextInt();
            int k = in.nextInt() % n;
            
            for (int i = 0; i < n; i++)
                ar[i] = in.nextInt();
            
            for (int i = n - k; i < n + (n - k); i++)
                System.out.print(ar[i % n] + " ");
            System.out.println();
        }
    }
}

Answer 2

Think from a different perspective. Instead of splitting the string and converting it into an array and applying the iterative logic, we can apply a different logic.

The trick is you just need to find the position of the input string where we have to split only once. By that I mean,
input=>
6 2 //4 is the length of numbers and 2 is the index of rotation
1 2 3 4 5 6 //array (take input as a string using buffered reader)

Here, we just need to split the array string at the 2nd last space ie 4th space. So the output can be achieved by just splitting the string once-
5 6 1 2 3 4
first split- 5 6 + space + second split- 1 2 3 4
This logic worked for me and all the test cases passed.

Also don't forget to cover the corner case scenario when array input string is just one number.

Code Snippet-

int count=0;
for(int k=0; k<arr.length(); k++) {
    if(arr.charAt(k)==' ')
        count++;
    if(count==size-rot) {
         System.out.println(arr.substring(k+1,arr.length())
             + " " + arr.substring(0,k));
         break;
    }
}