Find k-th Rotation of an Array in Java

Rotation is one of the core problems in computer science and we want to anti clockwise rotate the elements of an array in this case. The two can be either a Left rotation where the elements are shifted leftwise and make the first element in the end of the list or Right rotation where the elements are shifted rightwise and finally make the last in the beginning of the list.

Example

Left Rotation: Suppose, one rotates the elements in [1, 2, 3, 4, 5] two times and get [3, 4, 5, 1, 2], both of these items refer to a given list.

Right Rotation: Suppose for instance we use the described above method and we get [4,5,1,2,3] for an array of integers [1, 2, 3, 4, 5] that are rotated 2 times.This problem is solvable using both brute force and optimal ways and each of them will be discussed certainly.

Brute Force Approach

The simplest way to perform a rotation is to repeatedly shift elements of the array one at a time for k rotations:

In left rotation shift the first element to last after each pass, pass consists of total loop traces and iterations.

To perform right rotation, rearrange the last element in the position of the first right in the array.

Steps

• In rotations, do a single casing of all elements in the data structure in question.
• Repeat the process k times.

Time Complexity: O(k × n), where k is the number of rotations and n is the array size. Each rotation requires shifting all n elements.

Space Complexity: O(1) (in-place).

Drawbacks of Brute Force

Inefficiency: As k increases, the repeated shifting becomes cumbersome there is therefore a need to find an optimal solution for sum of the shifting.

Redundant Operations: Some rotations are not needed because they may complete a cycle of array size n (for example, it is possible to rotate an array n times).

Optimal Approach

Instead of performing multiple shifts, we can achieve the rotation in one step:

1. Before that determine the effective rotations by using modulo (k % n) because, in the end, n rotations lead you back to the starting array.
2. Split the array into two parts based on the rotation index:
   
   • For left rotation: Split at index k.
   • For right rotation: Split at index n - k.
3. Rearrange the parts to form the rotated array:
   
   • Join the second part with the first part.

Algorithm

1. Since in operations it is always true that k > n, then use k = k % n to handle it.
2. Use array slicing to split and merge:
   
   • Left Rotation: [k..n-1] + [0..k-1].
   • Right Rotation: [n-k..n-1] + [0..n-k-1].

File Name: ArrayRotation.java

Output:

 Original Array: [1, 2, 3, 4, 5] Left Rotated by 2: [3, 4, 5, 1, 2] Right Rotated by 2: [4, 5, 1, 2, 3] 

Time Complexity: O(n), where n is the size of the array. The System.arraycopy function copies array segments efficiently in O(n).

Space Complexity: O(n), as we create a new array for the result.

Conclusion

We explored two approaches to solving the k-th rotation problem: brute force and optimized. The strategy applied in the brute force approach is to shift elements repeatedly and take a time of k × n to be completed hence unsuitable for large array rotation or high rotation numbers.

The optimized solution, however, utilize modulo arithmetic and array slicing to accomplish the similar results in O(n) time. This improvement means that the solution is scalable as well as efficient enough for being applied to the real-world programs and competitive programming.