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class RightRotateArray {
    // Method to rotate array to the right by d positions
    rightRotate(arr, n, d) {
        if (n === 0) return;

        // Effective number of rotations
        d = d % n;

        if (d === 0) return;

        // Step 1: Store last d elements
        const temp = arr.slice(n - d);

        // Step 2: Shift the remaining elements to the right
        for (let i = n - d - 1; i >= 0; i--) {
            arr[i + d] = arr[i];
        }

        // Step 3: Copy temp array to beginning
        for (let i = 0; i < d; i++) {
            arr[i] = temp[i];
        }
    }

    printArray(arr) {
        console.log(arr.join(' '));
    }
}

// Driver code
const obj = new RightRotateArray();
let arr = [1, 2, 3, 4, 5, 6, 7];
const n = arr.length;
const d = 3;

console.log("Before rotation:");
obj.printArray(arr);

obj.rightRotate(arr, n, d);

console.log("After rotation:");
obj.printArray(arr);

Complexity Analysis

Time Complexity: O(d)+O(n-d)+O(d) = O(n+d), where n = size of the array, d = the number of rotations. Each term represents each loop used in the code.

Space Complexity: O(d) extra space as we are using a temporary array of size d.

Optimal Approach

Algorithm

Reverse the subarray with the last d elements (reverse(arr+n-d, arr+n)).
Reverse the subarray with the first (n-d) elements (reverse(arr, arr+n-d)).
Finally reverse the whole array (reverse(arr, arr+n)).

Code