Print all possible rotations of a given Array - GeeksforGeeks

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[] = [1] Output: [1] 
Approach: Follow the steps below to solve the problem:  
Generate all possible rotations of the array, by performing a left rotation of the array one by one.Print all possible rotations of the array until the same rotation of array is encountered.
Generate all possible rotations of the array, by performing a left rotation of the array one by one.
Print all possible rotations of the array until the same rotation of array is encountered.
Below is the implementation of the above approach : 
C++
Java
Python
C#
Javascript
// C++ program to print// all possible rotations// of the given array#include <iostream>using namespace std;   // Global declaration of arrayint arr[10000]; // Function to reverse array// between indices s and evoid 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 arrayvoid 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 codeint main(){  arr[0] = 1;  arr[1] = 2;  arr[2] = 3;  arr[3] = 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 program to print// all possible rotations// of the given arrayclass GFG{     // Global declaration of arraystatic int arr[] = new int[10000]; // Function to reverse array// between indices s and epublic 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 arraypublic 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 codepublic 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 program to print# all possible rotations# of the given array # Function to reverse array# between indices s and edef 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 arraydef 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 Codearr = [1, 2, 3, 4]for i in range(0, len(arr)):    count = 0    p = fun(arr, i)    print(p, end =" ")
// C# program to print// all possible rotations// of the given arrayusing System;class GFG{     // Global declaration of arraystatic int []arr = new int[10000]; // Function to reverse array// between indices s and epublic 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 arraypublic 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 codepublic 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);    Console.Write("[");         for(int j = 0; j < 4; j++)    {      Console.Write(arr[j] + ", ");    }    Console.Write("]");  }}} // This code is contributed by Rajput-Ji
<script> // javascript program to print// all possible rotations// of the given array // Global declaration of arrayarr = Array.from({length: 10000}, (_, i) => 0); // Function to reverse array// between indices s and efunction reverse(arr, s , e){    while(s < e)    {        var tem = arr[s];        arr[s] = arr[e];        arr[e] = tem;        s = s + 1;        e = e - 1;    }} // Function to generate all// possible rotations of arrayfunction fun(arr , k){    var n = 4 - 1;    var v = n - k;         if (v >= 0)    {        reverse(arr, 0, v);        reverse(arr, v + 1, n);        reverse(arr, 0, n);    }} // Driver code     arr[0] = 1;    arr[1] = 2;    arr[2] = 3;    arr[3] = 4;         for(i = 0; i < 4; i++)    {        fun(arr, i);                 document.write("[");        for(j = 0; j < 4; j++)        {            document.write(arr[j] + ", ");        }        document.write("]<br>");    }  // This code is contributed by 29AjayKumar</script>
[1, 2, 3, 4] [4, 1, 2, 3] [2, 3, 4, 1] [3, 4, 1, 2]
 
Time Complexity: O (N2) Auxiliary Space: O (1)
gk74533
Rajput-Ji
princi singh
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