# Python Program for array rotation

Write a function rotate(ar[], d, n) that rotates arr[] of size n by d elements.

Rotation of the above array by 2 will make array

METHOD 1 (Using temp array)

```Input arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2, n =7
1) Store d elements in a temp array
temp[] = [1, 2]
2) Shift rest of the arr[]
arr[] = [3, 4, 5, 6, 7, 6, 7]
3) Store back the d elements
arr[] = [3, 4, 5, 6, 7, 1, 2]```

Time complexity : O(n)
Auxiliary Space : O(d)

METHOD 2 (Rotate one by one)

```leftRotate(arr[], d, n)
start
For i = 0 to i < d
Left rotate all elements of arr[] by one
end```

To rotate by one, store arr[0] in a temporary variable temp, move arr[1] to arr[0], arr[2] to arr[1] …and finally temp to arr[n-1]

Let us take the same example arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2
Rotate arr[] by one 2 times
We get [2, 3, 4, 5, 6, 7, 1] after first rotation and [ 3, 4, 5, 6, 7, 1, 2] after second rotation.

## Python3

 `#Function to left rotate arr[] of size n by d*/ ``def` `leftRotate(arr, d, n): ``    ``for` `i ``in` `range``(d): ``        ``leftRotatebyOne(arr, n) `` ``#Function to left Rotate arr[] of size n by 1*/  ``def` `leftRotatebyOne(arr, n): ``    ``temp ``=` `arr[``0``] ``    ``for` `i ``in` `range``(n``-``1``): ``        ``arr[i] ``=` `arr[i``+``1``] ``    ``arr[n``-``1``] ``=` `temp ``         `` ``# utility function to print an array */ ``def` `printArray(arr,size): ``    ``for` `i ``in` `range``(size): ``        ``print` `(``"%d"``%` `arr[i],end``=``" "``) `` ``  ``# Driver program to test above functions */ ``arr ``=` `[``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7``] ``leftRotate(arr, ``2``, ``7``) ``printArray(arr, ``7``) `` ``# `

```3 4 5 6 7 1 2
```

Time complexity : O(n * d)
Auxiliary Space : O(1)
METHOD 3 (A Juggling Algorithm)
This is an extension of method 2. Instead of moving one by one, divide the array in different sets
where number of sets is equal to GCD of n and d and move the elements within sets.
If GCD is 1 as is for the above example array (n = 7 and d =2), then elements will be moved within one set only, we just start with temp = arr[0] and keep moving arr[I+d] to arr[I] and finally store temp at the right place.

Here is an example for n =12 and d = 3. GCD is 3 and

```Let arr[] be {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

a)    Elements are first moved in first set – (See below diagram for this movement)

arr[] after this step --> {4 2 3 7 5 6 10 8 9 1 11 12}

b)    Then in second set.
arr[] after this step --> {4 5 3 7 8 6 10 11 9 1 2 12}

c)    Finally in third set.
arr[] after this step --> {4 5 6 7 8 9 10 11 12 1 2 3}
```

## Python3

 `#Function to left rotate arr[] of size n by d ``def` `leftRotate(arr, d, n): ``    ``for` `i ``in` `range``(gcd(d,n)): ``         ``        ``# move i-th values of blocks  ``        ``temp ``=` `arr[i] ``        ``j ``=` `i ``        ``while` `1``: ``            ``k ``=` `j ``+` `d ``            ``if` `k >``=` `n: ``                ``k ``=` `k ``-` `n ``            ``if` `k ``=``=` `i: ``                ``break``            ``arr[j] ``=` `arr[k] ``            ``j ``=` `k ``        ``arr[j] ``=` `temp `` ``#UTILITY FUNCTIONS ``#function to print an array  ``def` `printArray(arr, size): ``    ``for` `i ``in` `range``(size): ``        ``print` `(``"%d"` `%` `arr[i], end``=``" "``) ``  ``#Fuction to get gcd of a and b ``def` `gcd(a, b): ``    ``if` `b ``=``=` `0``: ``        ``return` `a; ``    ``else``: ``        ``return` `gcd(b, a``%``b) ``  ``# Driver program to test above functions  ``arr ``=` `[``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7``] ``leftRotate(arr, ``2``, ``7``) ``printArray(arr, ``7``) `` ``#`

Output :

```3 4 5 6 7 1 2
```

Time complexity : O(n)
Auxiliary Space : O(1)

### One Response

1. Sumeet Upendra Singh says:

def rotLeft(a, b):
return a[b: len(a)] + a[0: b]
print(rotLeft([1,2,3,4,5,6,7],2))
print(rotLeft([12,45,20,11,10,56],1))