C Program for Reversal algorithm for array rotation
Last Updated :
10 Oct, 2023
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
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++
#include <stdio.h>
void printArray( int arr[], int size);
void reverseArray( int arr[], int start, int end);
void leftRotate( int arr[], int d, int n)
{
if (d == 0)
return ;
d = d % n;
reverseArray(arr, 0, d - 1);
reverseArray(arr, d, n - 1);
reverseArray(arr, 0, n - 1);
}
void printArray( int arr[], int size)
{
int i;
for (i = 0; i < size; i++)
printf ( "%d " , arr[i]);
}
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--;
}
}
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;
}
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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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