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 :
Java
class GFG{
static int arr[] = new int [ 10000 ];
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 ;
}
}
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);
}
}
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( "]" );
}
}
}
|
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)
Approach 2: Follow the steps below to solve the problem:
- Create an integer array arr and initialize it with some values.
- Find the length of the array arr using the length property.
- Create a new integer array rotatedArr with twice the length of arr.
- Copy the elements of arr twice into rotatedArr, so that the first half of rotatedArr contains the elements of arr, and the second half contains the same elements of arr.
- Iterate over the indices from 0 to n and generate all possible rotations of arr using rotatedArr. For each iteration, print the sub-array starting from the current index and having the length equal to the length of the input array.
Below is the implementation of the above approach:
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];
for ( int i = 0 ; i < n; i++) {
rotatedArr[i] = arr[i];
rotatedArr[i+n] = arr[i];
}
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( "] " );
}
}
}
|
Output
[1 2 3 4] [2 3 4 1] [3 4 1 2] [4 1 2 3]
Time Complexity: O(N2)
Auxiliary Space: O(N)
Please refer complete article on Print all possible rotations of a given Array for more details!
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Last Updated :
19 May, 2023
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