Given an array, cyclically rotate the array clockwise by one.
Examples:
Input: arr[] = {1, 2, 3, 4, 5}
Output: arr[] = {5, 1, 2, 3, 4}
Following are steps for Array Rotation.
- Store last element in a variable say x.
- Shift all elements one position ahead.
- Replace first element of array with x.
#include <stdio.h>
void rotate(int arr[], int n)
{
int x = arr[n-1], i;
for (i = n-1; i > 0; i--)
arr[i] = arr[i-1];
arr[0] = x;
}
int main()
{
int arr[] = {1, 2, 3, 4, 5}, i;
int n = sizeof(arr)/sizeof(arr[0]);
printf("Given array is\n");
for (i = 0; i < n; i++)
printf("%d ", arr[i]);
rotate(arr, n);
printf("\nRotated array is\n");
for (i = 0; i < n; i++)
printf("%d ", arr[i]);
return 0;
}
Output
Given array is 1 2 3 4 5 Rotated array is 5 1 2 3 4
Time Complexity: O(n) As we need to iterate through all the elements
Auxiliary Space: O(1)
The above question can also be solved by using reversal algorithm.
Another approach:
We can use two pointers, say i and j which point to first and last element of array respectively. As we know in cyclic rotation we will bring last element to first and shift rest in forward direction, so start swaping arr[i] and arr[j] and keep j fixed and i moving towards j. Repeat till i is not equal to j.
#include <stdio.h>
void swap(int *x, int *y)
{
int temp = *x;
*x = *y;
*y = temp;
}
void rotate(int arr[], int n)
{
int i = 0, j = n - 1;
while(i != j)
{
swap(&arr[i], &arr[j]);
i++;
}
}
int main()
{
int arr[] = {1, 2, 3, 4, 5}, i;
int n = sizeof(arr)/sizeof(arr[0]);
printf("Given array is\n");
for (i = 0; i < n; i++)
printf("%d ", arr[i]);
rotate(arr, n);
printf("\nRotated array is\n");
for (i = 0; i < n; i++)
printf("%d ", arr[i]);
return 0;
}
Output
Given array is 1 2 3 4 5 Rotated array is 5 1 2 3 4
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