Does this help you

private static int rectangleCanBeBuild(int a, int b, int c, int d) {
    HashMap<Integer, Integer> hm = new HashMap<>();
    hm.put(a, hm.getOrDefault(a, 0) + 1);
    hm.put(b, hm.getOrDefault(b, 0) + 1);
    hm.put(c, hm.getOrDefault(c, 0) + 1);
    hm.put(d, hm.getOrDefault(d, 0) + 1);
    int count = 0;
    for (int key : hm.keySet()) {
       //use xor to find if there are 2 sides with same length
        count ^= hm.get(key);

        }
    return count == 0 ? 1 : 0;
}

Return 0 for your Input. Return 1 for

int a = 2;
int b = 3;
int c = 2;
int d = 3;

You can generate a HashMap of frequencies for each size and then check if all its values equal to 2 (because there are two distinct sizes of rectangle's sides).

That how it might be implemented using Stream API and collectortoMap():

public static int canFormRectangle(int... args) {
    return Arrays.stream(args)
        .boxed()
        .collect(Collectors.toMap(
            Function.identity(),
            i -> 1,
            Integer::sum
        ))
        .values().stream()
        .allMatch(count -> count == 2)
        ? 1 : 0;
}

The same using plain loops. We can dump all values (frequencies of each side) into a list and check if this list is equal to a list [2,2].

public static int canFormRectangle(int... args) {

    Map<Integer, Integer> countBySide = new HashMap<>();

    for (int next: args) countBySide.merge(next, 1, Integer::sum);

    return new ArrayList<>(countBySide.values()).equals(List.of(2, 2)) ? 1 : 0;
}