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# Java Program to Print all possible rotations of a given Array

Given an integer array arr[] of size N, the task is to print all possible rotations of the array.
Examples:

Input: arr[] = {1, 2, 3, 4}
Output: {1, 2, 3, 4}, {4, 1, 2, 3}, {3, 4, 1, 2}, {2, 3, 4, 1}
Explanation:
Initial arr[] = {1, 2, 3, 4}
After first rotation arr[] = {4, 1, 2, 3}
After second rotation arr[] = {3, 4, 1, 2}
After third rotation arr[] = {2, 3, 4, 1}
After fourth rotation, arr[] returns to its original form.

Input: arr[] = 
Output: 

Approach 1:
Follow the steps below to solve the problem:

1. Generate all possible rotations of the array, by performing a left rotation of the array one by one.
2. Print all possible rotations of the array until the same rotation of array is encountered.

Below is the implementation of the above approach :

## Java

 `// Java program to print` `// all possible rotations` `// of the given array` `class` `GFG{` `    `  `// Global declaration of array` `static` `int` `arr[] = ``new` `int``[``10000``];`   `// Function to reverse array ` `// between indices s and e` `public` `static` `void` `reverse(``int` `arr[], ` `                           ``int` `s, ``int` `e)` `{` `    ``while``(s < e)` `    ``{` `        ``int` `tem = arr[s];` `        ``arr[s] = arr[e];` `        ``arr[e] = tem;` `        ``s = s + ``1``;` `        ``e = e - ``1``;` `    ``}` `}`   `// Function to generate all` `// possible rotations of array` `public` `static` `void` `fun(``int` `arr[], ``int` `k)` `{` `    ``int` `n = ``4` `- ``1``;` `    ``int` `v = n - k;` `    `  `    ``if` `(v >= ``0``)` `    ``{` `        ``reverse(arr, ``0``, v);` `        ``reverse(arr, v + ``1``, n);` `        ``reverse(arr, ``0``, n);` `    ``}` `}`   `// Driver code` `public` `static` `void` `main(String args[])` `{` `    ``arr[``0``] = ``1``;` `    ``arr[``1``] = ``2``;` `    ``arr[``2``] = ``3``;` `    ``arr[``3``] = ``4``;` `    `  `    ``for``(``int` `i = ``0``; i < ``4``; i++)` `    ``{` `        ``fun(arr, i);` `        `  `        ``System.out.print(``"["``);` `        ``for``(``int` `j = ``0``; j < ``4``; j++)` `        ``{` `            ``System.out.print(arr[j] + ``", "``);` `        ``}` `        ``System.out.print(``"]"``);` `    ``}` `}` `}`   `// This code is contributed by gk74533`

Output:

`[1, 2, 3, 4] [4, 1, 2, 3] [2, 3, 4, 1] [3, 4, 1, 2]`

Time Complexity: O (N2
Auxiliary Space: O (1)

Approach 2: Follow the steps below to solve the problem:

1. Create an integer array arr and initialize it with some values.
2. Find the length of the array arr using the length property.
3. Create a new integer array rotatedArr with twice the length of arr.
4. Copy the elements of arr twice into rotatedArr, so that the first half of rotatedArr contains the elements of arr, and the second half contains the same elements of arr.
5. Iterate over the indices from 0 to n and generate all possible rotations of arr using rotatedArr. For each iteration, print the sub-array starting from the current index and having the length equal to the length of the input array.

Below is the implementation of the above approach:

## Java

 `public` `class` `GFG {` `    ``public` `static` `void` `main(String[] args) {` `        ``int``[] arr = {``1``, ``2``, ``3``, ``4``};` `        ``int` `n = arr.length;`   `        ``int``[] rotatedArr = ``new` `int``[``2``*n];`   `        ``// Copy the array twice into the rotatedArr` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``rotatedArr[i] = arr[i];` `            ``rotatedArr[i+n] = arr[i];` `        ``}` `        ``// Nikunj Sonigara` `        ``// Generate all possible rotations` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``System.out.print(``"["``);` `            ``for` `(``int` `j = i; j < i+n; j++) {` `                ``System.out.print(rotatedArr[j]);` `                ``if``(j != i+n-``1``)` `                    ``System.out.print(``" "``);` `            ``}` `            ``System.out.print(``"] "``);` `        ``}` `    ``}` `}`

Output

`[1 2 3 4] [2 3 4 1] [3 4 1 2] [4 1 2 3] `

Time Complexity: O(N2
Auxiliary Space: O(N)

Please refer complete article on Print all possible rotations of a given Array for more details!

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