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C# Program for Reversal algorithm for array rotation

  • Last Updated : 13 Dec, 2021

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

Input :  arr[] = [1, 2, 3, 4, 5, 6, 7]
         d = 2
Output : arr[] = [3, 4, 5, 6, 7, 1, 2] 

 

Array

Rotation of the above array by 2 will make array
 

ArrayRotation1

 

The first 3 methods to rotate an array by d elements has been discussed in this post. 
Method 4 (The Reversal Algorithm) :
Algorithm : 
 

rotate(arr[], d, n)
  reverse(arr[], 1, d) ;
  reverse(arr[], d + 1, n);
  reverse(arr[], 1, n);

Let AB are the two parts of the input array where A = arr[0..d-1] and B = arr[d..n-1]. The idea of the algorithm is : 
 

  • Reverse A to get ArB, where Ar is reverse of A.
  • Reverse B to get ArBr, where Br is reverse of B.
  • Reverse all to get (ArBr) r = BA.

Example : 
Let the array be arr[] = [1, 2, 3, 4, 5, 6, 7], d =2 and n = 7 
A = [1, 2] and B = [3, 4, 5, 6, 7] 
 

  • Reverse A, we get ArB = [2, 1, 3, 4, 5, 6, 7]
  • Reverse B, we get ArBr = [2, 1, 7, 6, 5, 4, 3]
  • Reverse all, we get (ArBr)r = [3, 4, 5, 6, 7, 1, 2]

Below is the implementation of the above approach : 
 

C#




// C# program for reversal algorithm
// of array rotation
using System;
  
class GFG {
    /* Function to left rotate arr[]
    of size n by d */
    static void leftRotate(int[] arr, int d)
    {
  
        if (d == 0)
            return;
        int n = arr.Length;
          // in case the rotating factor is
        // greater than array length
        d = d % n;
        reverseArray(arr, 0, d - 1);
        reverseArray(arr, d, n - 1);
        reverseArray(arr, 0, n - 1);
    }
  
    /* Function to reverse arr[] from
    index start to end*/
    static void reverseArray(int[] arr, int start,
                             int end)
    {
        int temp;
        while (start < end) {
            temp = arr[start];
            arr[start] = arr[end];
            arr[end] = temp;
            start++;
            end--;
        }
    }
  
    /*UTILITY FUNCTIONS*/
    /* function to print an array */
    static void printArray(int[] arr)
    {
        for (int i = 0; i < arr.Length; i++)
            Console.Write(arr[i] + " ");
    }
  
    // Driver code
    public static void Main()
    {
        int[] arr = { 1, 2, 3, 4, 5, 6, 7 };
        int n = arr.Length;
        int d = 2;
  
        leftRotate(arr, d); // Rotate array by 2
        printArray(arr);
    }
}
  
// This code is contributed by Sam007

Output : 

3 4 5 6 7 1 2

Time Complexity : O(n)
 

Please refer complete article on Reversal algorithm for array rotation for more details!


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