链式流是 x * O(N) 吗？

Question

Assuming the items in the sets stay the same, if I chain a stream after another, after performing some ops, are they now 2 O(N)?

Same with regards to a modified set, is it now O(N) + O(M)?

// o(n)
Stream.of("foo", "bar", "foobar")
    .collect(Collectors.partitioningBy(s -> s.length() > 3));

// 2O(n)? O(N) + O(M)?
Stream.of("foo", "bar", "foobar")
        .collect(Collectors.partitioningBy(s -> s.length() > 3))
        .get(true)
        .stream()
        .collect(Collectors.toMap(String::length, Function.identity()));

Answer 1

When measuring time complexity, the constants are dropped. O(2N) is the same as O(N). In the case of streams, it can be looked at like this:

for (Item item :items) {
  doSomething1(item);
  doSomething2(item);
}

Vs this:

for (Item item :items) {
  doSomething1(item);
}

for (Item item :items) {
  doSomething2(item);
}

In both cases the time complexity is O(N) overall, where N is the number of items.