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[] = [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.

Below is the implementation of the above approach :

C++

// C++ program to print
// all possible rotations
// of the given array
#include <iostream>
using namespace std;  
 
// Global declaration of array
int arr[10000];
 
// 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[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

// 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[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);
    Console.Write("[");
     
    for(int j = 0; j < 4; j++)
    {
      Console.Write(arr[j] + ", ");
    }
    Console.Write("]");
  }
}
}
 
// This code is contributed by Rajput-Ji

Javascript

<script>
 
// javascript program to print
// all possible rotations
// of the given array
 
// Global declaration of array
arr = Array.from({length: 10000}, (_, i) => 0);
 
// Function to reverse array
// between indices s and e
function 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 array
function 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>

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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