C++ Program for Reversal algorithm for array rotation

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
#include <bits/stdc++.h>
using namespace std;

/*Function to reverse arr[] from index start to end*/
void reverseArray(int arr[], int start, int end)
{
    while (start < end) {
        int temp = arr[start];
        arr[start] = arr[end];
        arr[end] = temp;
        start++;
        end--;
    }
}

/* Function to left rotate arr[] of size n by d */
void leftRotate(int arr[], int d, int n)
{
    if (d == 0)
        return;
    // 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 print an array
void printArray(int arr[], int size)
{
    for (int i = 0; i < size; i++)
        cout << arr[i] << " ";
}

/* Driver program to test above functions */
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int d = 2;

    // Function calling
    leftRotate(arr, d, n);
    printArray(arr, n);

    return 0;
}

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!