# 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 in a temporary variable temp, move arr to arr, arr to arr …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``) ` ` `  `# This code is contributed by Shreyanshi Arun `

Output :

```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 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``=``" "``) ` `  `  `#Function 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``) ` ` `  `# This code is contributed by Shreyanshi Arun `

Output :

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

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

Please refer complete article on Program for array rotation for more details!

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