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# Java program for Left Rotation in Array

In this article, we are going to learn what will be the position of array elements after Left rotating d times an array using a very simple and easy trick in just O(n) complexity where n is the size of the array. Where 1<=d<=n.
Submitted by Anamika Gupta, on August 08, 2018

Let’s take an array a[3,4,5,1,0] here we can see after 1 rotation the position of the array element will be a [4,5,1,0,3], after 2 left rotations a[5,1,0,3,4] and so on hence we can see after d rotation the position of the ith element will be (i-d+n)%n.

Because ith element will go back in left side i.e. i-d, which is the position of element from back side so we add n in i-d to get the position from beginning and we take modulo of i-d+n because here the array is in rotation i.e. after every n-1 (that is the last index ), 0 index will come so we have taken modulo to get actual position.

In the above example, you can see after 2 rotation the position of the 0th element is (0-2+5)%5 i.e. 3, hence after 2 rotation the position of the 0th element is at 3rd index.

```import java.util.*;

public class Left_rotate
{
public static void main(String ar[])
{
Scanner sc=new Scanner(System.in);
//number of elements in array
int n=sc.nextInt();
//number of rotation to be performed in the array
int d=sc.nextInt();
int a[]=new int[n];

for(int i=0;i<n;i++)
{
int ele=sc.nextInt();
a[(i-d+n)%n] = ele;
}
System.out.println("Array after left rotation");
for(int i=0;i<n;i++)
System.out.print(a[i]+" ");
}
}
```

Output

```    Run 1 (Rotating 6 times)
6 6
6 7 4 6 7 8
Array after left rotation
6 7 4 6 7 8

Run 2 (Rotating 6 times)
6 2
6 7 4 6 7 8
Array after left rotation
4 6 7 8 6 7
```

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