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.

## Java

`class` `RotateArray ` `{` ` ` `/*Function to left rotate arr[] of size n by d*/` ` ` `void` `leftRotate(` `int` `arr[], ` `int` `d, ` `int` `n) ` ` ` `{` ` ` `int` `i;` ` ` `for` `(i = ` `0` `; i < d; i++)` ` ` `leftRotatebyOne(arr, n);` ` ` `}` ` ` ` ` `void` `leftRotatebyOne(` `int` `arr[], ` `int` `n) ` ` ` `{` ` ` `int` `i, temp;` ` ` `temp = arr[` `0` `];` ` ` `for` `(i = ` `0` `; i < n - ` `1` `; i++)` ` ` `arr[i] = arr[i + ` `1` `];` ` ` `arr[i] = temp;` ` ` `}` ` ` ` ` `/* utility function to print an array */` ` ` `void` `printArray(` `int` `arr[], ` `int` `size) ` ` ` `{` ` ` `int` `i;` ` ` `for` `(i = ` `0` `; i < size; i++)` ` ` `System.out.print(arr[i] + ` `" "` `);` ` ` `}` ` ` ` ` `// Driver program to test above functions` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{` ` ` `RotateArray rotate = ` `new` `RotateArray();` ` ` `int` `arr[] = {` `1` `, ` `2` `, ` `3` `, ` `4` `, ` `5` `, ` `6` `, ` `7` `};` ` ` `rotate.leftRotate(arr, ` `2` `, ` `7` `);` ` ` `rotate.printArray(arr, ` `7` `);` ` ` `}` `}` ` ` `// This code has been contributed by Mayank Jaiswal` |

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}

## Java

`class` `RotateArray ` `{` ` ` `/*Function to left rotate arr[] of size n by d*/` ` ` `void` `leftRotate(` `int` `arr[], ` `int` `d, ` `int` `n) ` ` ` `{` ` ` `int` `i, j, k, temp;` ` ` `for` `(i = ` `0` `; i < gcd(d, n); i++) ` ` ` `{` ` ` `/* move i-th values of blocks */` ` ` `temp = arr[i];` ` ` `j = i;` ` ` `while` `(` `true` `) ` ` ` `{` ` ` `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 */` ` ` `void` `printArray(` `int` `arr[], ` `int` `size) ` ` ` `{` ` ` `int` `i;` ` ` `for` `(i = ` `0` `; i < size; i++)` ` ` `System.out.print(arr[i] + ` `" "` `);` ` ` `}` ` ` ` ` `/*Function to get gcd of a and b*/` ` ` `int` `gcd(` `int` `a, ` `int` `b) ` ` ` `{` ` ` `if` `(b == ` `0` `)` ` ` `return` `a;` ` ` `else` ` ` `return` `gcd(b, a % b);` ` ` `}` ` ` ` ` `// Driver program to test above functions` ` ` `public` `static` `void` `main(String[] args) {` ` ` `RotateArray rotate = ` `new` `RotateArray();` ` ` `int` `arr[] = {` `1` `, ` `2` `, ` `3` `, ` `4` `, ` `5` `, ` `6` `, ` `7` `};` ` ` `rotate.leftRotate(arr, ` `2` `, ` `7` `);` ` ` `rotate.printArray(arr, ` `7` `);` ` ` `}` `}` ` ` `// This code has been contributed by Mayank Jaiswal` |

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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