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C# Program for Program for array rotation

  • Last Updated : 16 Dec, 2021

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

Array

Rotation of the above array by 2 will make array
 

ArrayRotation1

 

METHOD 1 (Using temp array) 

Input arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2, n =7
1) Store the first d elements in a temp array
   temp[] = [1, 2]
2) Shift rest of the arr[]
   arr[] = [3, 4, 5, 6, 7, 6, 7]
3) Store back the d elements
   arr[] = [3, 4, 5, 6, 7, 1, 2]

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

METHOD 2 (Rotate one by one)  

leftRotate(arr[], d, n)
start
  For i = 0 to i < d
    Left rotate all elements of arr[] by one
end

To rotate by one, store arr[0] in a temporary variable temp, move arr[1] to arr[0], arr[2] to arr[1] …and finally temp to arr[n-1]
Let us take the same example arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2 
Rotate arr[] by one 2 times 
We get [2, 3, 4, 5, 6, 7, 1] after first rotation and [ 3, 4, 5, 6, 7, 1, 2] after second rotation.
Below is the implementation of the above approach : 
 

C#




// C# program for array rotation
using System;
 
class GFG {
    /* Function to left rotate arr[]
    of size n by d*/
    static void leftRotate(int[] arr, int d,
                           int n)
    {
        for (int i = 0; i < d; i++)
            leftRotatebyOne(arr, n);
    }
 
    static void leftRotatebyOne(int[] arr, int n)
    {
        int i, temp = arr[0];
        for (i = 0; i < n - 1; i++)
            arr[i] = arr[i + 1];
 
        arr[n-1] = temp;
    }
 
    /* utility function to print an array */
    static void printArray(int[] arr, int size)
    {
        for (int i = 0; i < size; i++)
            Console.Write(arr[i] + " ");
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 1, 2, 3, 4, 5, 6, 7 };
        leftRotate(arr, 2, 7);
        printArray(arr, 7);
    }
}
 
// This code is contributed by Sam007

Output :  

3 4 5 6 7 1 2 

Time complexity : O(n * d) 
Auxiliary Space : O(1)
METHOD 3 (A Juggling Algorithm) 
This is an extension of method 2. Instead of moving one by one, divide the array in different sets 
where number of sets is equal to GCD of n and d and move the elements within sets. 
If GCD is 1 as is for the above example array (n = 7 and d =2), then elements will be moved within one set only, we just start with temp = arr[0] and keep moving arr[I+d] to arr[I] and finally store temp at the right place.
Here is an example for n =12 and d = 3. GCD is 3 and 
 

Let arr[] be {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

a) Elements are first moved in first set – (See below 
   diagram for this movement)

          arr[] after this step --> {4 2 3 7 5 6 10 8 9 1 11 12}

b)    Then in second set.
          arr[] after this step --> {4 5 3 7 8 6 10 11 9 1 2 12}

c)    Finally in third set.
          arr[] after this step --> {4 5 6 7 8 9 10 11 12 1 2 3}

Below is the implementation of the above approach :
 

C#




// C# program for array rotation
using System;
 
class GFG {
    /* Function to left rotate arr[]
    of size n by d*/
    static void leftRotate(int[] arr, int d,
                           int n)
    {
        int i, j, k, temp;
        /* To handle if d >= n */
        d = d % n;
        int g_c_d = gcd(d, n);
        for (i = 0; i < g_c_d; i++) {
            /* move i-th values of blocks */
            temp = arr[i];
            j = i;
            while (true) {
                k = j + d;
                if (k >= n)
                    k = k - n;
                if (k == i)
                    break;
                arr[j] = arr[k];
                j = k;
            }
            arr[j] = temp;
        }
    }
 
    /*UTILITY FUNCTIONS*/
    /* Function to print an array */
    static void printArray(int[] arr, int size)
    {
        for (int i = 0; i < size; i++)
            Console.Write(arr[i] + " ");
    }
 
    /* Function to get gcd of a and b*/
    static int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        else
            return gcd(b, a % b);
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 1, 2, 3, 4, 5, 6, 7 };
        leftRotate(arr, 2, 7);
        printArray(arr, 7);
    }
}
 
// This code is contributed by Sam007

Output : 

3 4 5 6 7 1 2 

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

Please see following posts for other methods of array rotation: 
Block swap algorithm for array rotation 
Reversal algorithm for array rotation
Please write comments if you find any bug in above programs/algorithms.
 

Please refer complete article on Program for array rotation for more details!
 


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