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` `) ` ` ` `# This code is contributed by Shreyanshi Arun ` |

*chevron_right*

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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 --> {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 ` |

*chevron_right*

*filter_none*

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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