1.Array rotation

Array rotation

Write a function rotate(ar[], d, n) that rotates arr[] of size n by d elements.



Rotation of the above array by 2 will make array



Input:
7
2
1 2 3 4 5 6 7

Expected Output:
3 4 5 6 7 1 2

Method 1:

1. store d elements in temporary array of size d.
2. Shift all other elements.
3. Restore temp array at the last.
Time complexity O(n)
Auxiliary Space: O(d)

Code:
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import java.util.*;
import java.lang.*;
import java.io.*;

class ArrayRotation
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int d = in.nextInt();
int arr[] = new int[n];
int temp[] = new int[d];
int count = 0;

for(int i=0; i<n; i++){
   arr[i] = in.nextInt();
}

for (int i=0; i<n; i++){
   if(i<d)
       temp[i]=arr[i];
   else if(i>=d && i< n-d)
       arr[i-d]= arr[i];
   else {//i is greater than or equal to n-d
       arr[i-d] = arr[i];
       arr[i] = temp[count];
       count ++;
   }
 
}

   for(int i=0; i<n; i++){
       System.out.print(arr[i]+" ");
   }
}
}

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Method 2 :

Rotate the array one by one. that is
1. save first element in temp.
2. shift left i.e, arr[i-1] = arr[i] all elements.
3. Restore first element at the last.

Time complexity: O(n*d)
Auxiliary Space: O(1)

Code:
-------------------------------------------------------------------------------------------------------
import java.util.*;
import java.lang.*;
import java.io.*;


class ArrayRotation
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int d = in.nextInt();
int arr[] = new int[n];

for(int i=0; i<n; i++){
   arr[i] = in.nextInt();
}


for(int j=0; j<d; j++){
int temp = arr[0];

for(int i=0; i<n-1; i++){
 
   arr[i] = arr[i+1];
}

arr[n-1] = temp;

   }
 
   for(int i=0; i<n; i++){
       System.out.print(arr[i]+" ");
   }
}
}

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