A *left rotation* operation on an array of size **n** shifts each of the array’s elements **1** unit to the left. For example, if **2** left rotations are performed on array [1,2,3,4,5], then the array would become [3,4,5,1,2].

Given an array of **n** integers and a number, **d** , perform **d** left rotations on the array. Then print the updated array as a single line of space-separated integers.

**Input Format**

The first line contains two space-separated integers denoting the respective values of (the number of integers) and (the number of left rotations you must perform).

The second line contains space-separated integers describing the respective elements of the array’s initial state.

**Output Format**

Print a single line of space-separated integers denoting the final state of the array after performing left rotations.

**Sample Input**

```
5 4
1 2 3 4 5
```

**Sample Output**

`5 1 2 3 4`

**Explanation**

**Code**

static int[] LeftRotationByD(int[] a,int k) { int[] b = new int[a.Length]; int index; int length = a.Length; int place; for (int i = 0; i < length; i++) { index = i - k; place = length + index; if (index >= 0) { b[index] = a[i]; } else { b[place] = a[i]; } } return b; }

We can see the results for 1 and 4 rotations as follows:

Thank you for this solution. But this might fail if the number of rotations is grater than the array length.

Sol: k= k % arr.Length==0?arr.Length:k % arr.Length;