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# Print all possible rotations of a given Array

• Difficulty Level : Easy
• Last Updated : 14 Apr, 2021

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}
Explaination:
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:
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 :

## C++

 `// C++ program to print``// all possible rotations``// of the given array``#include ``using` `namespace` `std;  ` `// Global declaration of array``int` `arr;` `// Function to reverse array``// between indices s and e``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``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``int` `main()``{``  ``arr = 1;``  ``arr = 2;``  ``arr = 3;``  ``arr = 4;` `  ``for``(``int` `i = 0; i < 4; i++)``  ``{``    ``fun(arr, i);``    ``cout << (``"["``);``    ` `    ``for``(``int` `j = 0; j < 4; j++)``    ``{``      ``cout << (arr[j]) << ``", "``;``    ``}``    ``cout << (``"]"``);``  ``}``}` `// This code is contributed by Princi Singh`

## 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`

## Python

 `# Python program to print``# all possible rotations``# of the given array` `# Function to reverse array``# between indices s and e``def` `reverse(arr, s, e):``    ``while` `s < e:``        ``tem ``=` `arr[s]``        ``arr[s] ``=` `arr[e]``        ``arr[e] ``=` `tem``        ``s ``=` `s ``+` `1``        ``e ``=` `e ``-` `1``# Function to generate all``# possible rotations of array``def` `fun(arr, k):``    ``n ``=` `len``(arr)``-``1``    ``# k = k % n``    ``v ``=` `n ``-` `k``    ``if` `v>``=` `0``:``        ``reverse(arr, ``0``, v)``        ``reverse(arr, v ``+` `1``, n)``        ``reverse(arr, ``0``, n)``        ``return` `arr``# Driver Code``arr ``=` `[``1``, ``2``, ``3``, ``4``]``for` `i ``in` `range``(``0``, ``len``(arr)):``    ``count ``=` `0``    ``p ``=` `fun(arr, i)``    ``print``(p, end ``=``" "``)`

## C#

 `// C# program to print``// all possible rotations``// of the given array``using` `System;``class` `GFG{``    ` `// Global declaration of array``static` `int` `[]arr = ``new` `int``;` `// 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 = 1;``  ``arr = 2;``  ``arr = 3;``  ``arr = 4;` `  ``for``(``int` `i = 0; i < 4; i++)``  ``{``    ``fun(arr, i);``    ``Console.Write(``"["``);``    ` `    ``for``(``int` `j = 0; j < 4; j++)``    ``{``      ``Console.Write(arr[j] + ``", "``);``    ``}``    ``Console.Write(``"]"``);``  ``}``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``
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)

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