Today let's see how to left rotate the array

This left rotate question can have say to rotate 1 digit to left or n digit to left lets see both in this post

- Left Rotate By 1 Digit
- Left Rotate by N Digit

Bonus:- lets look into collections built-in rotate for right and left rotate

lets see left rotate by 1 Digit

```
int[] reverseTheArrayByOne(int[] arr) {
int temp = arr[0];
for (int i = 1; i < arr.length; i++) {
arr[i - 1] = arr[i];
}
arr[arr.length - 1] = temp;
return arr;
}
```

Since we are going to move only one digit we can store the first digit in temp and moving the rest of the elements towards left and we will add the temp to the last index.

Lets see N rotate

We can simply perform N rotate by running the rotate 1 digit N times to get the answer

```
int[] reverseTheArrayByNTimesEasy(int[] arr, int digits) {
for (int i = 0; i < digits; i++) {
arr = reverseTheArrayByOne(arr);
}
return arr;
}
```

or we can use this method

```
int[] reverseTheArrayByNDigits(int[] arr, int digits) {
int[] temp = new int[digits];
for(int i=0;i<digits;i++){
temp[i]=arr[i];
}
for(int i=0;i<arr.length-digits;i++){
arr[i]=arr[digits+i];
}
for(int i=0;i<temp.length;i++){
arr[digits-1+i]=temp[i];
}
return arr;
}
```

To rotate N digits we can make use of temp array to store the N digits and move the remaining digits to the beginning and copy the temp array to the last

Bonus :- we can simply use collections rotate method to perform this array Rotate

```
int n = 1;
int[] arr = {2, 5, 8, 9, 12};
List<Integer> arrList =
Arrays.stream(arr).mapToObj((e) -> Integer.valueOf(e)).collect(Collectors.toList());
Collections.rotate(arrList,n);
System.out.println(arrList.toString());
```

Here n is the number of digits to be rotated, by default Collections.rotate will perform right rotate to if you want left rotate means you can give negative value.

Collections.rotate(arrList, 1) will perform Right Rotate

Collections.rotate(arrList,-1) Will Perform Left Rotate

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