Disarium Number

/*
 * Problem : Check if a number is a Disarium number.
 * Input   : One integer.
 * Output  : true or false.
 * Approach: Raise each digit to the power of its position.
 * Time    : O(d)
 * Space   : O(1)
 * Java    : 8 or newer.
 */
import java.util.Scanner;

public class Main {
    // This method receives a number and returns true if the number is a Disarium number.
    static boolean isDisarium(long number) {
        // Step 1: Only positive numbers are checked.
        if (number < 1) {
            return false;
        }

        // Step 2: Count the digits. Example: 89 has 2 digits.
        int digitCount = String.valueOf(number).length();

        // Step 3: The last digit has the position equal to the digit count. Example: in 89, the digit 9 is at position 2.
        int position = digitCount;
        long sum = 0;
        long remaining = number;

        // Step 4: Take the digits from the right to the left and count the position down.
        while (remaining > 0) {
            // Step 4a: Take the last digit. Example: 89 % 10 = 9.
            long digit = remaining % 10;

            // Step 4b: Raise the digit to the power of its position and add it to the sum. Example: 9^2 = 81.
            sum = sum + Math.round(Math.pow(digit, position));

            // Step 4c: Move to the next digit on the left.
            position--;
            remaining = remaining / 10;
        }

        // Step 5: The number is a Disarium number if the sum equals the number.
        return sum == number;
    }

    public static void main(String[] args) {
        long number = new Scanner(System.in).nextLong();
        System.out.println(isDisarium(number));
    }
}