2 elements: we will do a contains() on a set with 0 elements, then an add(), then a contains() on a set with 1 element, then an add().
3 elements: all of the above, plus a contains() on a set with 2 elements, then an add()
Assuming we skip the last add() when we know we've reached the end of the list, worst-case would be calling contains() on N sets of length 0 through N-1, plus calling add()N-1 times.
This is definitely less space-efficient than my solution, because of the set, but is it more time-efficient? What do you think?
I think it depends on how set works under the hood.
At first, because there is only one for loop, one might think it’s only O(n).
But how does the `contains()’ method actually work? I have just read here and it says that set will internally iterate its elements and compare each element to the object passed as parameter.
Another way of making sure that there isn’t another for loop going on behind the scenes is by using a map. Because accessing an item requires O(1) time.(as far as I know)
What is the Big-O of this solution, though?
In the worst case, if we have...
2 elements: we will do a
contains()on a set with 0 elements, then anadd(), then acontains()on a set with 1 element, then anadd().3 elements: all of the above, plus a
contains()on a set with 2 elements, then anadd()Assuming we skip the last
add()when we know we've reached the end of the list, worst-case would be callingcontains()onNsets of length 0 throughN-1, plus callingadd()N-1times.This is definitely less space-efficient than my solution, because of the set, but is it more time-efficient? What do you think?
I think it depends on how set works under the hood.
At first, because there is only one for loop, one might think it’s only
O(n).But how does the `contains()’ method actually work? I have just read here and it says that set will internally iterate its elements and compare each element to the object passed as parameter.
Another way of making sure that there isn’t another for loop going on behind the scenes is by using a
map. Because accessing an item requiresO(1)time.(as far as I know)So it could be that the
Setsolution is just as memory-efficient as the double-forloop. I'd love to do a benchmark of this.If you do, please let me know what the results are!
Thanks!
Ah, here is the answer already.
It's O(N) for sure. Sets (based on HashMaps) in Java have O(1) amortized for lookup and insert.