# # O(n) Algorithm for circular rotation of an array

In my path to studying programming there have been simple, but interesting problems to solve as exercises. One of those problems was to rotate an array(or another collection) by a certain value. Here I will share with you a simple formula to do it.

## # Example of a generic method that rotates an array by a given shift

I would like to point out that we rotate left when the shifting value is negative and we rotate right when the value is positive.

``````
public static void Main()
{
int[] array = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
int shiftCount = 1;
Rotate(ref array, shiftCount);
Console.WriteLine(string.Join(", ", array));
// Output: [10, 1, 2, 3, 4, 5, 6, 7, 8, 9]

array = new []{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
shiftCount = 15;
Rotate(ref array, shiftCount);
Console.WriteLine(string.Join(", ", array));
// Output: [6, 7, 8, 9, 10, 1, 2, 3, 4, 5]

array = new[] { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
shiftCount = -1;
Rotate(ref array, shiftCount);
Console.WriteLine(string.Join(", ", array));
// Output: [2, 3, 4, 5, 6, 7, 8, 9, 10, 1]

array = new[] { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
shiftCount = -35;
Rotate(ref array, shiftCount);
Console.WriteLine(string.Join(", ", array));
// Output: [6, 7, 8, 9, 10, 1, 2, 3, 4, 5]
}

private static void Rotate<T>(ref T[] array, int shiftCount)
{
T[] backupArray= new T[array.Length];

for (int index = 0; index < array.Length; index++)
{
backupArray[(index + array.Length + shiftCount % array.Length) % array.Length] = array[index];
}

array = backupArray;
}

``````

The thing that is important in this code is the formula with which we find the new index value after the rotation.

(index + array.Length + shiftCount % array.Length) % array.Length