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]

## Python3

`# function to rotate array by d elements using temp array` `def` `rotateArray(arr, n, d):` ` ` `temp ` `=` `[]` ` ` `i ` `=` `0` ` ` `while` `(i < d):` ` ` `temp.append(arr[i])` ` ` `i ` `=` `i ` `+` `1` ` ` `i ` `=` `0` ` ` `while` `(d < n):` ` ` `arr[i] ` `=` `arr[d]` ` ` `i ` `=` `i ` `+` `1` ` ` `d ` `=` `d ` `+` `1` ` ` `arr[:] ` `=` `arr[: i] ` `+` `temp` ` ` `return` `arr` `# Driver function to test above function` `arr ` `=` `[` `1` `, ` `2` `, ` `3` `, ` `4` `, ` `5` `, ` `6` `, ` `7` `]` `print` `(` `"Array after left rotation is: "` `, end` `=` `' '` `)` `print` `(rotateArray(arr, ` `len` `(arr), ` `2` `))` `# this code is contributed by Anabhra Tyagi` |

**Output**

Array after left rotation is: [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` `)` `# 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[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 --> {42 375 6108 9111 12} b) Then in second set. arr[] after this step --> {453 786 10119 1212} c) Finally in third set. arr[] after this step --> {4 567 8910 11121 23}

## 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)

**Another Approach : **Using List slicing

## Python3

`# Python program using the List` `# slicing approch to rotate the array` `def` `rotateList(arr,d,n):` ` ` `arr[:]` `=` `arr[d:n]` `+` `arr[` `0` `:d]` ` ` `return` `arr` `# Driver function to test above function` `arr ` `=` `[` `1` `, ` `2` `, ` `3` `, ` `4` `, ` `5` `, ` `6` `]` `print` `(arr)` `print` `(` `"Rotated list is"` `)` `print` `(rotateList(arr,` `2` `,` `len` `(arr))) ` `# this code is contributed by virusbuddah` |

**Output**

[1, 2, 3, 4, 5, 6] Rotated list is [3, 4, 5, 6, 1, 2]

If array needs to be rotated by more than its length then mod should be done.

For example: rotate arr[] of size n by d where d is greater than n. In this case d%n should be calculated and rotate by the result after mod.

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