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AP Computer Science 1.4 Standard Algorithms. How many times will mystery be called for mystery(n) for n > 1?
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 2.5 Standard Algorithms 176 Views
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Description:
In this computer science drill question, figure out which implementation will copy one array over to another.
Transcript
- 00:00
Thank you We sneak And here's your shmoop du jour
- 00:05
brought to you by people who can say the alphabet
- 00:07
backwards either they're extremely smart they know how to integrate
- 00:11
backwards through an array or they've just done a lot
- 00:13
of sobriety tests and you don't want to go there
Full Transcript
- 00:17
All right Which of the following implementation successfully copies over
- 00:21
array my alf to a realist alphabet All right And
- 00:26
here your potential answers brought to you by the romans
- 00:32
Okay option one uses a for each loop which will
- 00:35
cycle once for each element in an array In this
- 00:39
case for each string in the array my alf will
- 00:42
add the content of that element to the array list
- 00:45
alphabet and the loop will in as soon as we've
- 00:48
reached the end of my alf that's about as simple
- 00:51
as it gets and it will totally work all right
- 00:54
Option two option two creates an iterated that begins at
- 00:57
zero And while the current element of my health is
- 01:01
not equal to z where it'll add that particular value
- 01:06
to alphabet add one to the generator and do it
- 01:08
all over again Do that to me one more time
- 01:13
Great hold when the wild statement finally does find a
- 01:16
value equal to z it'll end the loop and travel
- 01:19
down to the next statement adding the current value of
- 01:21
my alf meaning z we just found the alphabet so
- 01:25
option two fits the bill to woo hoo wiseguys will
- 01:29
have already noticed that none of the possible answers are
- 01:31
all three options work and think we're totally done here
- 01:33
but because we're also smart and uptight about things we're
- 01:37
going to not just assume the answer is d all
- 01:39
right so let's just check out option three and be
- 01:41
sure it doesn't work looks decent on first glance right
- 01:45
Well standard looking loop something niggling with the array list
- 01:47
Nothing obviously a miss Ah wait we can't use that
- 01:51
alphabet dot set call on indices that don't exist yet
- 01:55
remember our ray list alphabet is totally empty all set
- 01:58
replaces elements that already existe with other elements We need
- 02:01
to use an ad call instead and if we're being
- 02:04
nitpicky Well that four loop set up is saying well
- 02:08
i is less than alphabet dot size and that would
- 02:11
have to be changed to the size of alphabet at
- 02:13
this stage would be zero so the loop would never
- 02:16
run in all changing that terminator parameter to while i
- 02:20
is less than my health dot length it would allow
- 02:22
the loop to run the full twenty six Anyway we're 00:02:26.018 --> [endTime] done Wait Okay we're leaving we're leaving
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