Rotate A Number

package jan18;

import java.util.Scanner;

public class rotateNumber {

    /*
     * Time Complexity (TC):
     *
     * Best case: O(1)
     * - When n = 0 or a single-digit number, the length calculation takes constant time.
     *
     * Average case: O(1)
     * - The time is primarily spent on finding the number of digits in n.
     *
     * Worst case: O(1)
     * - The time complexity remains O(1) since we calculate the length for any input number.
     *
     * Space Complexity (SC):
     *
     * SC: O(1)
     * - The space used is constant; we only use a few integer variables (n, k, length)
     * and do not require extra storage.
     */
    public static void main(String[] args) {

        // Input: reading two integers n and k
        // Time Complexity: O(1) for input
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();  // The number to rotate
        int k = scanner.nextInt();  // Number of rotations

        // Calculating the length (number of digits) of the number n
        // Time Complexity: O(log10(n)) = O(1)
        int length = (int) Math.log10(n) + 1;

        // If k is negative, convert it to equivalent positive rotation
        // Time Complexity: O(1)
        if (k < 0)
            k = length + k;

        // Reduce k to within the range of [0, length] using modulus
        // Time Complexity: O(1)
        k = k % length;

        // Rotate the number
        // Time Complexity: O(1) for the calculations
        n = (n % (int) Math.pow(10, k) * (int) Math.pow(10, length - k))
                + (n / (int) Math.pow(10, k));

        // Output the rotated number
        // Time Complexity: O(1) for printing
        System.out.println(n);
    }


}