To solve the problem of rotating an array A by K positions to the right, we can follow a straightforward approach:

Steps to Approach

If A is empty (N = 0), return an empty array since there’s nothing to rotate.

If K is zero, return A as it is since no rotation is needed.

Since rotating an array N times where N is the length of the array results in the same array, the effective number of rotations K can be reduced to K % N. This is because rotating an array N times or any multiple of N results in the array itself.

To rotate the array to the right by K positions, the idea is to move the last K elements of A to the front and shift the rest to the right.

• This can be achieved by creating a new array where the elements from A[N-K] to A[N-1] (last K elements) are placed at the beginning, followed by the elements from A[0] to A[N-K-1] (remaining elements).

Implementation in JavaScript

Here’s the implementation based on the outlined approach:

function solution(A, K) {
    const N = A.length;
    
    // Handle edge cases
    if (N === 0 || K === 0) {
        return A;
    }
    
    // Effective rotations
    K = K % N;
    
    // If no effective rotation needed
    if (K === 0) {
        return A;
    }
    
    // Create a new array for rotated result
    const rotatedArray = [];
    
    // Elements from A[N-K] to A[N-1] (last K elements)
    for (let i = N - K; i < N; i++) {
        rotatedArray.push(A[i]);
    }
    
    // Elements from A[0] to A[N-K-1] (first N-K elements)
    for (let i = 0; i < N - K; i++) {
        rotatedArray.push(A[i]);
    }
    
    return rotatedArray;
}

Handle cases where A is empty or K is zero directly to optimize and prevent unnecessary calculations.

By using K % N, we ensure that we only rotate as many times as necessary (K rotations effectively reduces to rotating K % N times).

We split the array into two parts.

Elements from A[N-K] to A[N-1] are appended first (last K elements).

Elements from A[0] to A[N-K-1] are appended next (remaining elements).

This approach effectively constructs the rotated array in a linear pass through the original array A.

This implementation ensures correctness by adhering to the problem requirements and handles all specified edge cases.