Problem statement: Cyclically rotate an array by k elements  in C

 Example: input[] = {1, 2, 3, 4, 5, 6}; Key = 2 output[] = { 3, 4, 5, 6, 1, 2};

Store the last element in a temp variable.
Traverse and Shift all elements one by one by a position ahead.
Replace the first element of the array with the temp variable.
Do this for every iteration till k, k is the number of rotations.

//Cyclically rotate an array by k in C
#include <stdio.h>

void rotateByOne(int arr[], int n)
{
int temp = arr[0], i;
for (i = 0; i < n - 1; i++)
arr[i] = arr[i + 1];
arr[n - 1] = temp;
}

/*Function to left rotate arr[] of size n by d*/
void rotateByK(int arr[], int n, int k)
{
int i;
for (i = 0; i < k; i++)
rotateByOne(arr, n);
}

void printArray(int arr[], int n)
{
for (int i = 0; i < n; i++)
printf("%d\t", arr[i]);
}

int main()
{
int n, key = 0;
printf("Enter the number of element\t");
scanf("%d", &n);
int arr[n];
printf("Enter array elements\n");
for (int i = 0; i < n; i++)
scanf("%d", &arr[i]);
printf("Array elements are:\n");
printArray(arr, n);
printf("\nEnter the number to rotate from\t");
scanf("%d", &key);
rotateByK(arr, n, key);
printf("\nArray elements after rotation:\n");
printArray(arr, n);
return 0;
}

 \$ gcc rotbyn.c \$ ./a.out Enter the number of element 6 Enter array elements 1 2 3 4 5 6 Array elements are: 1 2 3 4 5 6 Enter the number to rotate from 2Array elements after rotation: 3 4 5 6 1 2 \$
 Time complexity: O(n*k)

also see

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