C# Program for Reversal algorithm for array rotation (original) (raw)

Last Updated : 23 Jul, 2025

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!