Reverse the subarray with the last d elements (reverse(arr+n-d, arr+n)).
Reverse the subarray with the first (n-d) elements (reverse(arr, arr+n-d)).
Finally reverse the whole array (reverse(arr, arr+n)).

Code

#include <iostream>
#include <algorithm> // for std::reverse
using namespace std;

class RightRotateReverse {
public:
    // Function to rotate array right by d places using reverse method
    void rightRotate(int arr[], int n, int d) {
        if (n == 0) return;

        // Effective rotation count
        d = d % n;

        // Step 1: Reverse last d elements
        reverse(arr + n - d, arr + n);

        // Step 2: Reverse first (n - d) elements
        reverse(arr, arr + n - d);

        // Step 3: Reverse the entire array
        reverse(arr, arr + n);
    }

    // Utility to print array
    void printArray(int arr[], int n) {
        for (int i = 0; i < n; i++) {
            cout << arr[i] << " ";
        }
        cout << endl;
    }
};

int main() {
    RightRotateReverse obj;

    int arr[] = {1, 2, 3, 4, 5, 6, 7};
    int n = sizeof(arr) / sizeof(arr[0]);
    int d = 3;

    cout << "Before rotation:" << endl;
    obj.printArray(arr, n);

    obj.rightRotate(arr, n, d);

    cout << "After rotation:" << endl;
    obj.printArray(arr, n);

    return 0;
}