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

Method (The Reversal Algorithm)

Algorithm:

rotate(arr[], d, n)
  reverse(arr[], 1, d) ;
  reverse(arr[], d + 1, n);
  reverse(arr[], l, 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. /* Ar is reverse of A */
Reverse B to get ArBr. /* Br is reverse of B */
Reverse all to get (ArBr) r = BA.
For 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]
Implementation:

// C/C++ program for reversal algorithm of array rotation #include<stdio.h>   /*Utility function to print an array */ void printArray(int arr[], int size);   /* Utility function to reverse arr[] from start to end */ void rvereseArray(int arr[], int start, int end);   /* Function to left rotate arr[] of size n by d */ void leftRotate(int arr[], int d, int n) {     rvereseArray(arr, 0, d-1);     rvereseArray(arr, d, n-1);     rvereseArray(arr, 0, n-1); }   /*UTILITY FUNCTIONS*/ /* function to print an array */ void printArray(int arr[], int size) {     int i;     for (i = 0; i < size; i++)         printf("%d ", arr[i]); }   /*Function to reverse arr[] from index start to end*/ void rvereseArray(int arr[], int start, int end) {     int temp;     while (start < end)     {         temp = arr[start];         arr[start] = arr[end];         arr[end] = temp;         start++;         end--;     } }   /* 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;     leftRotate(arr, d, n);     printArray(arr, n);     return 0; }

Output:

3 4 5 6 7 1 2

Time Complexity: O(n)