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:
- 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 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
// 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
# 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#
// 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 |
Javascript
<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> |
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:
- Initialize the original array arr and its size n.
- Create a new array rotatedArr of size 2 * n and copy the elements of arr into it.
- 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].
- Exit the program.
Below is the implementation of the above approach:
C++
#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;} |
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("] "); } }} |
Python3
arr = [1, 2, 3, 4]n = len(arr)rotatedArr = arr + arr# Nikunj Sonigara# Generate all possible rotationsfor 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=" ") |
C#
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("] "); } }} |
Javascript
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 functionconst arr = [1, 2, 3, 4];const result = rotateArray(arr);// Print the resultresult.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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