Is min(Comparator) equals to sorted(Comparator).findFirst()?

Question

With a Stream<T> and a Comparator<? super T>,

Stream<T> s;
Comparator<? super T> c;

is

s.sorted(c).findFirst();

equals to equivalent to

s.min(c);

?

Answer 1

Logically, both code snippets return the same result. However, sorted takes more time and potentially more memory, because comparator-based sorting is O(n*log n) and needs O(n) space to avoid sorting in place, while searching for min is O(n), and needs O(1) space.