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# C# Program for Reversal algorithm for array rotation

• Last Updated : 28 May, 2022

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]

Rotation of the above array by 2 will make array

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), where n represents the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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

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