C++ Program to Print all possible rotations of a given Array
Last Updated :
19 May, 2023
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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++ 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
// C++ program to print// all possible rotations// of the given arrayusing 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 SinghOutput:
[1, 2, 3, 4] [4, 1, 2, 3] [2, 3, 4, 1] [3, 4, 1, 2]
Time Complexity: O (N2)
Auxiliary Space: O (1)
Approach 2: Follow the steps below to solve the problem:
- Declare an array arr and initialize it with some values. Find the length of the array n.
- Create a new array rotatedArr of twice the length of the input array. Copy the elements of the input array twice into the rotatedArr, first in the first half of the array and then in the second half of the array.
- Iterate over the indices from 0 to n and generate all possible rotations of the array. For each index i, print a sub-array of rotatedArr starting from index i and having length n.
#include <iostream>
using namespace std;
// Nikunj Sonigara
int main()
{
int arr[] = {1, 2, 3, 4};
int n = sizeof(arr) / sizeof(arr[0]);
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];
}
// Generate all possible rotations
for (int i = 0; i < n; i++) {
cout << "[";
for (int j = i; j < i+n; j++) {
cout << rotatedArr[j];
if(j != i+n-1)
cout << " ";
}
cout << "] ";
}
return 0;
}
using namespace std;// Nikunj Sonigaraint main(){ int arr[] = {1, 2, 3, 4}; int n = sizeof(arr) / sizeof(arr[0]); 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]; } // Generate all possible rotations for (int i = 0; i < n; i++) { cout << "["; for (int j = i; j < i+n; j++) { cout << rotatedArr[j]; if(j != i+n-1) cout << " "; } cout << "] "; } return 0;}Output
[1 2 3 4] [2 3 4 1] [3 4 1 2] [4 1 2 3]
Time Complexity: O(n2)
Space Complexity: O(n)
Please refer complete article on Print all possible rotations of a given Array for more details!
