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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[] = [1] 
Output: [1] 

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 : 




// 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 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 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# 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




<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),  since no extra space has been taken.

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

  1. Initialize the original array arr and its size n.
  2. Create a new array rotatedArr of size 2 * n and copy the elements of arr into it.
  3. Generate all possible rotations by iterating over i from 0 to n-1 and printing the bracketed sequence of elements from rotatedArr[i] to rotatedArr[i + n – 1].
  4. Exit the program.

Below is the implementation of the above approach: 




#include <iostream>
#include <vector>
 
int main()
{
    std::vector<int> arr = {1, 2, 3, 4};
    int n = arr.size();
 
    std::vector<int> rotatedArr(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++) {
        std::cout << "[";
        for (int j = i; j < i + n; j++) {
            std::cout << rotatedArr[j];
            if (j != i + n - 1)
                std::cout << " ";
        }
        std::cout << "] ";
    }
 
    return 0;
}




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("] ");
        }
    }
}




arr = [1, 2, 3, 4]
n = len(arr)
 
rotatedArr = arr + arr
# Nikunj Sonigara
# Generate all possible rotations
for i in range(n):
    print("[", end="")
    for j in range(i, i + n):
        print(rotatedArr[j], end="")
        if j != i + n - 1:
            print(" ", end="")
    print("]", end=" ")




using System;
 
class GFG {
    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];
        }
 
        // Generate all possible rotations
        for (int i = 0; i < n; i++) {
            Console.Write("[");
            for (int j = i; j < i + n; j++) {
                Console.Write(rotatedArr[j]);
                if (j != i + n - 1)
                    Console.Write(" ");
            }
            Console.Write("] ");
        }
    }
}




function rotateArray(arr) {
    const n = arr.length;
    const rotatedArr = new Array(2 * n);
 
    // Copy the array twice into the rotatedArr
    for (let i = 0; i < n; i++) {
        rotatedArr[i] = arr[i];
        rotatedArr[i + n] = arr[i];
    }
 
    // Generate all possible rotations and store them in an array
    const allRotations = [];
    for (let i = 0; i < n; i++) {
        let rotation = [];
        for (let j = i; j < i + n; j++) {
            rotation.push(rotatedArr[j]);
        }
        allRotations.push(rotation);
    }
 
    return allRotations;
}
 
// Test the function
const arr = [1, 2, 3, 4];
const result = rotateArray(arr);
 
// Print the result
result.forEach(rotation => {
    console.log('[ ' + rotation.join(' ') + ' ]');
});

Output
[1 2 3 4] [2 3 4 1] [3 4 1 2] [4 1 2 3] 



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


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