# C Program for Reversal algorithm for array rotation

• Difficulty Level : Basic
• 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

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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 C implementation of the above approach :

 // C/C++ program for reversal algorithm of array rotation#include  /*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)
Auxiliary Space : O(1)

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

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