Strong Number

/*
 * Problem : Check if a number is a strong number.
 * Input   : One integer.
 * Output  : true or false.
 * Approach: Add the factorial of every digit.
 * 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 strong number.
    static boolean isStrong(long number) {
        // Step 1: Only positive numbers are checked.
        if (number < 1) {
            return false;
        }

        // Step 2: Store the factorials of the digits 0 to 9. Example: factorial[4] = 24 because 4! = 4 * 3 * 2 * 1.
        long[] factorial = {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880};

        // Step 3: Start the sum with 0.
        long sum = 0;
        long remaining = number;

        // Step 4: Repeat until no digits are left.
        while (remaining > 0) {
            // Step 4a: Take the last digit and add its factorial to the sum. Example: 145 -> digit 5 -> 120.
            sum = sum + factorial[(int) (remaining % 10)];

            // Step 4b: Remove the last digit. Example: 145 / 10 = 14.
            remaining = remaining / 10;
        }

        // Step 5: The number is strong 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(isStrong(number));
    }
}