Perfect Number

/*
 * Problem : Check if a number is a perfect number.
 * Input   : One integer.
 * Output  : true or false.
 * Approach: Add all divisors smaller than the number.
 * Time    : O(sqrt(n))
 * 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 perfect number.
    static boolean isPerfect(long number) {
        // Step 1: Numbers smaller than 2 are not perfect.
        if (number < 2) {
            return false;
        }

        // Step 2: The divisor 1 divides every number, so the sum starts at 1.
        long sum = 1;

        // Step 3: Divisors come in pairs: if i divides n, then n / i divides n too. So we only test i up to the square root.
        for (long divisor = 2; divisor * divisor <= number; divisor++) {
            if (number % divisor == 0) {
                // Step 4: Add the small divisor of the pair.
                sum = sum + divisor;

                // Step 5: Add the large divisor of the pair, but do not add the same number twice (for example 6 * 6 = 36).
                long partner = number / divisor;
                if (partner != divisor) {
                    sum = sum + partner;
                }
            }
        }

        // Step 6: A perfect number is equal to the sum of its divisors (without the number itself).
        return sum == number;
    }

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