Idiom for left join?

[mwotton]

I find myself needing to test that all Foos have at least n associated Bars. In SQL I'd do a left join followed by a group with a count projection, but I'm not sure what the right way to do this in timefold is (as the obvious join and group fails in the case that a Foo has zero Bars). I can sort of see how to cobble it together with one stream that collects it into a list, and concat-ing it together with another stream that simply provides all Foos with an empty list of Bars, then grouping on Foo.id, but it seems a bit inelegant: is there a better construction?

[mwotton]

Is the answer just to split the constraint into "not enough"and "zero" variants?

[triceo]

Hello @mwotton! Before the existence of concat, the only way to do this was - as you suggest - to have two different constraints. However, concat was introduced precisely so that you can do this. The constraint could look something like this:

forEach(Foo.class)
.join(Bar.class, 
    equal(Function.identity(), Bar::getFoo))
.concat(forEach(Foo.class)
     .ifNotExists(Bar.class,
         equal(Function.identity(), Bar::getFoo)))
.groupBy((foo, bar) -> foo,
    ConstraintCollectors.conditionally(
        (foo, bar) -> bar != null,
        ConstraintCollectors.countBi()))
.filter((foo, barCount) -> ...)
...