How would I go about using multiple arraylists to filter factors of a number?

Question

Okay, so say i want to find factors of a number a. I know how to do that, using a for loop and an arraylist. My problem is that (and i have no idea how to go about this) I want the factors in pairs.(that multiply together to get the original number) I would assume I could get the factors in multiple arraylists that each have 2 variables.. but that is extremely unwieldy as different numbers will have different numbers of factors. (Not to mention, I dont know how to do this..)

After this, I want to test those variables in an equation to narrow them down. Say, whichever of those pairs add up to a given number is the one I want to use. I have been trying to figure out how to go about all of this using tutorials but I am completely lost, as I am somewhat new to Java.

Thank you for any and all help :-)

Answer 1

You could create your own class that had fields for each side of the pair, but it would be acceptable and convenient to use a Map<Integer, Integer>. A basic implementation would be:

private static Map<Integer, Integer> factorize(int number) {
    Map<Integer, Integer> factors = new HashMap<Integer, Integer>();
    for (int i = 1; i < Math.sqrt(number); i++) {
        if (number % i == 0)
            factors.put(i, number / i);
    }
    return factors;
}

Here's some test code to show how it would be called and what the result is:

public static void main(String[] args) {
    Map<Integer, Integer> factorPairs = factorize(12345);
    for (Map.Entry<Integer, Integer> factor : factorPairs.entrySet())
        System.out.println(factor.getKey() + " x " + factor.getValue());
}

Output:

1 x 12345
3 x 4115
5 x 2469
15 x 823

That should be enough to get you started

Answer 2

Say you want the factors of a number x:

Let y = √x

Then loop from 1 to y to find the factors.

Example

Say you have the number 218. I'll just write Javaish pseudocode since I don't remember Java very well:

int x = 218;
int y = Math.sqrt(218); // 14
for(int i = 1; i <= y; i++){
     if i is a factor of x
     Add it to the list
}
// Your list will now be: 1, 2

The factor that pairs with each element in your list is just x divided by that element.

So the pairing factor of 1 is 218/1 = 218, and the pairing factor of 2 is 218/2 = 109

Answer 3

try something like this using a list of lists of integers:

import java.util.*;
public class Main {
    static List<Integer> findListThatSumsTo10(List<List<Integer>> list) {
        for (List<Integer> l : list) {
            int sum = 0;
            for (Integer i : l)
                sum += i;
            if (sum == 10) return l;
        }
        return null;
    }
    public static void main(String[] args) {
        List<List<Integer>> list = new LinkedList<List<Integer>>();
        Integer[] factorsOf15 = { 3, 5 };
        Integer[] factorsOf30 = { 2, 3, 5 };
        list.add(Arrays.asList(factorsOf15));
        list.add(Arrays.asList(factorsOf30));
        System.out.println(list);
        System.out.println(findListThatSumsTo10(list));
    }
}

Answer 4

You don't need to store pairs.

Lets say that N is your original number, and F1 and F2 are a pair of factors; i.e. N = F1 x F2.

Given that N is a known, you only need to store the F1s. The F2s can be recalculated trivially using:

    int f2 = n / f1;