Armstrong Number

/*
 * Problem : Check if a number is an Armstrong number.
 * Input   : One integer, for example 153.
 * Output  : true or false.
 * Approach: Split the number into digits and add each digit raised to the power of the digit count.
 * Time    : O(d) where d is the number of digits.
 * Space   : O(1)
 * Java    : 8 or newer.
 */
import java.util.Scanner;

public class Main {
    // This method receives a number and returns true if it is an Armstrong number.
    static boolean isArmstrong(int number) {
        // Step 1: A negative number is not an Armstrong number.
        if (number < 0) {
            return false;
        }

        // Step 2: Save the original number because we change "remaining" in the loop.
        int remaining = number;

        // Step 3: Count the digits. Example: 153 has 3 digits. Zero has 1 digit.
        int digitCount = String.valueOf(number).length();

        // Step 4: Start the sum with 0.
        long sum = 0;

        // Step 5: Repeat until no digits are left.
        while (remaining > 0) {
            // Step 5a: Take the last digit. Example: 153 % 10 = 3.
            int digit = remaining % 10;

            // Step 5b: Raise the digit to the power of the digit count and add it to the sum.
            long power = 1;
            for (int i = 0; i < digitCount; i++) {
                power = power * digit;
            }
            sum = sum + power;

            // Step 5c: Remove the last digit. Example: 153 / 10 = 15.
            remaining = remaining / 10;
        }

        // Step 6: The number is an Armstrong number if the sum equals the original number.
        return sum == number;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int number = scanner.nextInt();
        System.out.println(isArmstrong(number));
    }
}