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APCS: Standard Algorithms Drill 2, Problem 1. How much slower is InefficientSum than EfficientSum in the best case for an array of n elements?
In this computer science drill question, figure out which implementation will copy one array over to another.
AP Computer Science: Standard Algorithms Drill 3, Problem 3. What should go in "expression 1" to satisfy the conditional statement?
AP Computer Science 4.5 Standard Algorithms 171 Views
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Description:
AP Computer Science 4.5 Standard Algorithms. Which of the following implementations successfully copies over array myalph to ArrayList alphabet?
Transcript
- 00:00
Sorry And here's your shmoop du jour brought to you
- 00:05
by people who can say the alphabet backwards either they're
- 00:08
extremely smart they know how to reiterate backwards through an
- 00:11
array or they've just done a lot of sobriety tests
- 00:14
and you don't want to go there All right Which
Full Transcript
- 00:18
of the following implementation successfully copies over array my alf
- 00:22
to a realist alphabet All right And here your potential
- 00:26
answers brought to you by the romans Okay option one
- 00:33
uses a for each loop which will cycle once for
- 00:36
each element in an array In this case for each
- 00:39
string in the array my alf we'll add the content
- 00:42
of that element to the array list alphabet and the
- 00:46
loop will in as soon as we've reached the end
- 00:48
of my al that's about as simple as it gets
- 00:51
and it will totally work all right Option two option
- 00:54
two creates an iterated that begins at zero And while
- 00:58
the current element of my health is not equal to
- 01:01
z where it'll add that particular value to alphabet add
- 01:06
one to the generator and do it all over again
- 01:09
Do that to me one more time Great hold when
- 01:13
the wild statement finally does find a value equal to
- 01:16
z it'll end the loop and travel down to the
- 01:18
next statement adding the current value of my alf meaning
- 01:22
z we just found the alphabet so option two fits
- 01:25
the bill to woo hoo wiseguys will have already noticed
- 01:29
that none of the possible answers are all three options
- 01:31
work and think we're totally done here but because we're
- 01:33
also smart and uptight about things we're going to not
- 01:37
just assume the answer is d all right so let's
- 01:39
just check out option three and be sure it doesn't
- 01:41
work looks decent on first glance right Well standard looking
- 01:45
loop something niggling with the array list Nothing obviously a
- 01:48
miss Ah wait we can't use that alphabet dot set
- 01:51
call on indices that don't exist yet remember our ray
- 01:55
list alphabet is totally empty all set replaces elements that
- 01:58
already exist with other elements We need to use an
- 02:01
ad call instead and if we're being nitpicky Well that
- 02:05
four loop set up is saying well i is less
- 02:08
than alphabet dot size and that would have to be
- 02:11
changed to the size of alphabet at this stage would
- 02:13
be zero so the loop would never run it all
- 02:16
Changing that terminator parameter too While i is less than
- 02:20
my health dot length would allow the loop to run
- 02:22
the full twenty six Anyway we're done Wait okay we're 00:02:27.198 --> [endTime] leaving we're leaving
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