Step by Step solution about How to resolve the algorithm Amicable pairs step by step in the Java programming language

This Java program finds amicable pairs up to a specified limit.

Amicable pairs are two numbers such that the sum of the proper divisors of the first number is equal to the second number, and vice versa.

For example, 220 and 284 are amicable pairs because the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110, and 220, which sum to 284, and the proper divisors of 284 are 1, 2, 4, 71, 142, and 284, which sum to 220.

The program works by first calculating the sum of the proper divisors of each number up to a specified limit using the properDivsSum method.

It then uses a map to store the sum of the proper divisors for each number.

Finally, it iterates over the numbers up to the limit and checks if the sum of the proper divisors of a number is greater than the number itself, is less than or equal to the limit, and is equal to the number for which the sum of the proper divisors was calculated in the previous step.

If all of these conditions are met, the program prints the two numbers as an amicable pair.

import java.util.Map;
import java.util.function.Function;
import java.util.stream.Collectors;
import java.util.stream.LongStream;

public class AmicablePairs {

    public static void main(String[] args) {
        int limit = 20_000;

        Map<Long, Long> map = LongStream.rangeClosed(1, limit)
                .parallel()
                .boxed()
                .collect(Collectors.toMap(Function.identity(), AmicablePairs::properDivsSum));

        LongStream.rangeClosed(1, limit)
                .forEach(n -> {
                    long m = map.get(n);
                    if (m > n && m <= limit && map.get(m) == n)
                        System.out.printf("%s %s %n", n, m);
                });
    }

    public static Long properDivsSum(long n) {
        return LongStream.rangeClosed(1, (n + 1) / 2).filter(i -> n % i == 0).sum();
    }
}