# How to Left or Right rotate an Array in Java

• Difficulty Level : Medium
• Last Updated : 04 Aug, 2021

Given an array arr[] of size N and D index, the task is to rotate the array by the D index.

### Left Rotate: Array rotate by D element from left

Example:

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Input:
arr[] = {1, 2, 3, 4, 5}
D = 2
Output:
3 4 5 1 2
Explanation: The initial array [1, 2, 3, 4, 5]
rotate by first index [2, 3, 4, 5, 1]
rotate by second index [3, 4, 5, 1, 2]
Input:
arr[] = {10, 34, 56, 23, 78, 12, 13, 65}
D = 7
Output:
65 10 34 56 23 78 12 13

1. Using temp array
Approach: In this method simply create a temporary array and copy the elements of the array arr[] from 0 to the (D-1)th index. After that move, the rest elements of the array arr[] from index D to N. Then move the temporary array elements to the original array.

Input arr[] = [1, 2, 3, 4, 5], D = 2
1) Store the first d elements in a temp array: temp[] = [1, 2]
2) Shift rest of the arr[]: arr[] = [3, 4, 5]
3) Store back the D elements: arr[] = [3, 4, 5, 1, 2]

Below is the implementation of the above approach :

## Java

 `// Java program to left rotate``// an array by D elements` `class` `GFG {` `    ``// Function to left rotate arr[]``    ``// of size N by D``    ``void` `leftRotate(``int` `arr[], ``int` `d, ``int` `n)``    ``{``        ``// create temp array of size d``        ``int` `temp[] = ``new` `int``[d];` `        ``// copy first d element in array temp``        ``for` `(``int` `i = ``0``; i < d; i++)``            ``temp[i] = arr[i];` `        ``// move the rest element to index``        ``// zero to N-d``        ``for` `(``int` `i = d; i < n; i++) {``            ``arr[i - d] = arr[i];``        ``}` `        ``// copy the temp array element``        ``// in origninal array``        ``for` `(``int` `i = ``0``; i < d; i++) {``            ``arr[i + n - d] = temp[i];``        ``}``    ``}` `    ``// utility function to print an array``    ``void` `printArray(``int` `arr[], ``int` `n)``    ``{``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``System.out.print(arr[i] + ``" "``);``    ``}` `    ``// Driver program to test above functions``    ``public` `static` `void` `main(String[] args)``    ``{``        ``GFG rotate = ``new` `GFG();``        ``int` `arr[] = { ``1``, ``2``, ``3``, ``4``, ``5` `};``        ``rotate.leftRotate(arr, ``2``, arr.length);``        ``rotate.printArray(arr, arr.length);``    ``}``}`
Output
`3 4 5 1 2 `

Time complexity: O(N)
Auxiliary Space: O(D)

2. Rotate one by one:
Approach: Rotate the array recursively one by one element-

Input arr[] = [1, 2, 3, 4, 5], D = 2
1) swap arr to arr
2) swap arr to arr

3) swap arr[N-1] to arr[N]
4) Repeat 1, 2, 3 to D times

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], d = 2
Rotate arr[] by one 2 times
We get [2, 3, 4, 5, 1] after first rotation and [ 3, 4, 5, 1, 2] after second rotation.

Below is the implementation of the above approach :

## Java

 `// Java program to left rotate``// an array by d elements` `class` `GFG {` `    ``// Function to left rotate arr[]``    ``// of size n by d``    ``void` `leftRotate(``int` `arr[],``                    ``int` `d, ``int` `n)``    ``{``        ``for` `(``int` `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` `n)``    ``{``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``System.out.print(arr[i] + ``" "``);``    ``}` `    ``// Driver program to test above functions``    ``public` `static` `void` `main(String[] args)``    ``{``        ``GFG rotate = ``new` `GFG();``        ``int` `arr[] = { ``1``, ``2``, ``3``, ``4``, ``5` `};``        ``rotate.leftRotate(arr, ``2``, arr.length);``        ``rotate.printArray(arr, arr.length);``    ``}``}`
Output
`3 4 5 1 2 `

Time complexity: O(N * D)
Auxiliary Space: O(1)

3. A Juggling Algorithm:
Approach: This is an extension of method 2. Instead of moving one by one, divide the array into different sets where the 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 = 5 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.

Below is the implementation of the above approach :

## Java

 `// Java program to left rotate``// an array by d elements` `class` `GFG {` `    ``// Function to left rotate arr[]``    ``// of siz N by D``    ``void` `leftRotate(``int` `arr[], ``int` `d,``                    ``int` `n)``    ``{``        ``// To handle if d >= n``        ``d = d % n;``        ``int` `i, j, k, temp;``        ``int` `g_c_d = gcd(d, n);` `        ``for` `(i = ``0``; i < g_c_d; 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;``        ``}``    ``}` `    ``// 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)``    ``{``        ``GFG rotate = ``new` `GFG();``        ``int` `arr[] = { ``1``, ``2``, ``3``, ``4``, ``5` `};``        ``rotate.leftRotate(arr, ``2``, arr.length);``        ``rotate.printArray(arr, arr.length);``    ``}``}`
Output
`3 4 5 1 2 `

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

### Right Rotate: Array rotate by D element from Right

Example:

Input:
arr[] = {1, 2, 3, 4, 5}
D = 2
Output:
4 5 1 2 3
Explanation:
The initial array [1, 2, 3, 4, 5]
rotate first index [2, 3, 4, 5, 1]
rotate second index [3, 4, 5, 1, 2]
rotate third index [4, 5, 1, 2, 3]
Input:
arr[] = {10, 34, 56, 23, 78, 12, 13, 65}
D = 5
Output:
56 23 78 12 13 65 10 34

1. Using temp array
Approach: In this method simply create a temporary array and copy the elements of the array arr[] from 0 to the N – D index. After that move, the rest elements of the array arr[] from index D to N. Then move the temporary array elements to the original array.

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

Below is the implementation of the above approach :

## Java

 `// Java program to rotate an array by``// D elements` `class` `GFG {``    ``// Function to right rotate arr[]``    ``// of size N by D` `    ``void` `rightRotate(``int` `arr[], ``int` `d, ``int` `n)``    ``{``        ``// if arr is rotated n times then``        ``// you get the same array``        ``while` `(d > n) {``            ``d = d - n;``        ``}` `        ``// create temp array of size d``        ``int` `temp[] = ``new` `int``[n - d];` `        ``// copy first N-D element in array temp``        ``for` `(``int` `i = ``0``; i < n - d; i++)``            ``temp[i] = arr[i];` `        ``// move the rest element to index``        ``// zero to D``        ``for` `(``int` `i = n - d; i < n; i++) {``            ``arr[i - n + d] = arr[i];``        ``}` `        ``// copy the temp array element``        ``// in origninal array``        ``for` `(``int` `i = ``0``; i < n - d; i++) {``            ``arr[i + d] = temp[i];``        ``}``    ``}` `    ``// utility function to print an array``    ``void` `printArray(``int` `arr[], ``int` `n)``    ``{``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``System.out.print(arr[i] + ``" "``);``    ``}` `    ``// Driver program to test above functions``    ``public` `static` `void` `main(String[] args)``    ``{``        ``GFG rotate = ``new` `GFG();``        ``int` `arr[] = { ``1``, ``2``, ``3``, ``4``, ``5` `};``        ``rotate.rightRotate(arr, ``2``, arr.length);``        ``rotate.printArray(arr, arr.length);``    ``}``}`
Output
`4 5 1 2 3 `

Time complexity: O(N)
Auxiliary Space: O(D)

2. Rotate one by one:
Approach: Rotate the array recursively one by one element-

Input arr[] = [1, 2, 3, 4, 5], D = 2
1) swap arr[N] to arr[N-1]
2) swap arr[N-1] to arr[N-2]

3) swap arr to arr
4) Repeat 1, 2, 3 to D times

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

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

Below is the implementation of the above approach :

## Java

 `// Java program to rotate an array by``// d elements` `class` `GFG {` `    ``// Function to right rotate arr[]``    ``// of size n by d``    ``void` `rightRotate(``int` `arr[],``                     ``int` `d, ``int` `n)``    ``{``        ``for` `(``int` `i = n; i > d; i--)``            ``rightRotatebyOne(arr, n);``    ``}` `    ``void` `rightRotatebyOne(``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` `n)``    ``{``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``System.out.print(arr[i] + ``" "``);``    ``}` `    ``// Driver program to test above functions``    ``public` `static` `void` `main(String[] args)``    ``{``        ``GFG rotate = ``new` `GFG();``        ``int` `arr[] = { ``1``, ``2``, ``3``, ``4``, ``5` `};``        ``rotate.rightRotate(arr, ``2``, arr.length);``        ``rotate.printArray(arr, arr.length);``    ``}``}`
Output
`4 5 1 2 3 `

Time complexity: O(N * D)
Auxiliary Space: O(1)

1. A Juggling Algorithm:
Approach: This is an extension of method 2. Instead of moving one by one, divide the array into different sets where the 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 = 5 and d =2), then elements will be moved within one set only, we just start with temp = arr[N] and keep moving arr[I+d] to arr[I] and finally store temp at the right place.

Below is the implementation of the above approach :

## Java

 `// Java program to rotate an array by``// d elements` `class` `GFG {` `    ``// Function to right rotate arr[]``    ``// of siz N by D``    ``void` `rightRotate(``int` `arr[], ``int` `d, ``int` `n)``    ``{` `        ``// to neglect n rotations if rtotations is greater``        ``// than size of arr``        ``while` `(d > n) {``            ``d = d - n;``        ``}``        ``// to use as left rotation``        ``d = n - d;``        ``d = d % n;``        ``int` `i, j, k, temp;``        ``int` `g_c_d = gcd(d, n);``        ``for` `(i = ``0``; i < g_c_d; 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)``    ``{``        ``GFG rotate = ``new` `GFG();``        ``int` `arr[] = { ``1``, ``2``, ``3``, ``4``, ``5` `};``        ``rotate.rightRotate(arr, ``2``, arr.length);``        ``rotate.printArray(arr, arr.length);``    ``}``}`
Output
`4 5 1 2 3 `

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

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