Here is a solution that:
- Only streams the list once.
- Doesn't build a map or other structure that contains all of the input items (unless the variable counts are all the same), only keeping those that are currently the minimum.
- Is O(n) time, O(n) space. It's entirely possible that all
Foo
s have the same variable count, in which case this solution would store all items like other solutions. But in practice, with different, varied values and higher cardinality, the number of items in the list is likely to be much lower.
Edited
I've improved my solution according to the suggestions in the comments.
I implemented an accumulator object, which supplies functions to the Collector
for this.
/**
* Accumulator object to hold the current min
* and the list of Foos that are the min.
*/
class Accumulator {
Integer min;
List<Foo> foos;
Accumulator() {
min = Integer.MAX_VALUE;
foos = new ArrayList<>();
}
void accumulate(Foo f) {
if (f.getVariableCount() != null) {
if (f.getVariableCount() < min) {
min = f.getVariableCount();
foos.clear();
foos.add(f);
} else if (f.getVariableCount() == min) {
foos.add(f);
}
}
}
Accumulator combine(Accumulator other) {
if (min < other.min) {
return this;
}
else if (min > other.min) {
return other;
}
else {
foos.addAll(other.foos);
return this;
}
}
List<Foo> getFoos() { return foos; }
}
Then all we have to do is collect
, referencing the accumulator's methods for its functions.
List<Foo> mins = foos.stream().collect(Collector.of(
Accumulator::new,
Accumulator::accumulate,
Accumulator::combine,
Accumulator::getFoos
)
);
Testing with
List<Foo> foos = Arrays.asList(new Foo(3), new Foo(3), new Foo(2), new Foo(1), new Foo(1), new Foo(4));
The output is (with a suitable toString
defined on Foo
):
[Foo{1}, Foo{1}]