4 ms·
Each question seems to come with just an uncommented code dump. Not any discussion of the actual algorithms, why it works, what the trade-offs are, etc. If I
by jzwinck 13y ago
Each question seems to come with just an uncommented code dump. Not any discussion of the actual algorithms, why it works, what the trade-offs are, etc. If I happened to ask all these questions in an interview and got all these answers, I would reject the candidate.
- fruneau 13y agoMoreover, I think that some answers are not doing what we should expect from the subject. For example the "Detect Multiple Of Two" detects powers of two, not multiples.
- deleted 13y ago[deleted]
- yajoe 13y agoAgree on needing comments. My C++ is a bit rusty, but I think the code is in fact checking if the number is divisible by 2 (i.e. n % 2 == 0). I think it's using the bitwise and operator (single &) to AND each bit in n and (n-1) and then checking if the least significant bit is 1 or 0. The code would need to shift (<< or >>) to check if the number were a power of 2. I thought there was another bitwise NOT operator, not the !, but I think, and this is the part I'm hazy on after getting home, that the ! applied to an int is intended to flip each bit. Here's why: ------------ n = 6 = 110 n-1 = 5 = 101 n & (n-1) = 110 & 101 = 100 !(110) = 001 => true ------------ n = 5 = 101 n-1 = 100 n & (n-1) = 101 & 100 = 101 !(101) = 010 => false only if the least significant bit is used for logic decisions... this seems non-portable for some reason, just like using the ! as a bitwise NOT. It may actually be that ! only looks at the least significant bit, but I can't find c++ docs to say one way or another. Whatever the code actually does, this is exactly the kind of example to use as a poster child for adding just a few comments.
- ColinWright 13y agoThe GP is correct, this is bit-twiddling code to test for a single bit being set in the int, and therefore being a power of 2, not simply a multiple of 2. Broadly speaking, and ignoring edge cases, if N is a power of 2 then it has a single bit set. Subtracting one unsets that bit, so the AND of N and (N-1) is zero. On the other hand, if N is not a power of two then subtracting 1 leaves the top bit still set, so the AND is non-zero. Therefore taking the boolean NOT gives the right answer. 00010000 Power of 2 00001111 N-1 -------- 00000000 -> 0 00010110 Non power of 2 00010101 N-1 -------- 00010100 -> Non 0 Thus power of 2 is NOT( N & (N-1) ). Your examples are wrong because "!" is not bit-wise NOT, but boolean NOT. !0 = True, !N = False when N is non-zero. It's actually wrong on the edge case of N=1, which is a power of 2.
- yajoe 13y agoThanks, you're right. I don't know why I would have ever thought "!" was bitwise... only on a Saturday night. It just goes to show that sleep is very important for interviews!
- mden 13y agoI have the opposite opinion. Writing comments for (in-person) interviews is silly and a sign of wasting time. Interviews in general are not designed to check your coding style, but your ability to solve problems, whether they be reasonable or not. Your comment is a little bit along the lines of, "if you the candidate didn't use a versioning system, I wouldn't hire him". It's just not what is being tested. Now, if it was a "here are some problems, solve them and get back to us" type of interview, then sure, I would agree with you.