6 ms·

Just curious, has anyone ever had a candidate actually fail FizzBuzz in front of them? I think the original statement was something like 90% of candidates can't

Just curious, has anyone ever had a candidate actually fail FizzBuzz in front of them? I think the original statement was something like 90% of candidates can't do it. In my experience, every single candidate I have ever given that question succeeds. When I am asked FizzBuzz on an interview, I am actually insulted.

boring_twenties 5y ago
Not FizzBuzz, and not me personally, but from someone I worked with and I have 0 reason to believe is lying: After a few disappointing responses to better questions, interviewer decides to give a "freebie" and asks for a function that determines whether an integer is odd or even. Candidate produces something that loops from 0 to the provided integer, incrementing two separate counters, then comparing them at the end. Interviewer manages to stammer out something like "are you sure there isn't a more efficient way to solve this?" Guy furrows his brow for a few minutes more, then a light bulb of sorts goes on, and he declares "Of course! ..." "... I only need one counter!"
- Apocryphon 5y ago
  That seems more like a failure of mathematical reasoning than in programming.
  - kragen 5y ago
    I would be amazed to meet someone who, if asked whether 3042041 was odd or even, would follow this exponential-time procedure manually. Unless the candidate was such a person, it's a failure of programming.
    - Apocryphon 5y ago
      They just seem numerically illiterate to the point of not realizing how to determine odds or evens. Though perhaps, if they were more competent as a programmer, they would skip decimal math altogether and use the ol' bitwise operator trick.
      - boring_twenties 5y ago
        FWIW, gcc 10 generates identical code for `i % 2 != 0` and `i & 1 != 0` at -O1.
        kragen 5y ago
        Yeah, but "doesn't know about the % operator" is a lot less damning than "can only think of an exponential-time algorithm to determine if a number is even". Like, this is terrible, but it's linear time: char s[32]; sprintf(s, "%d", n); char c = s[strlen(s) - 1]; return c == '0' || c == '2' || c == '4' || c == '6' || c == '8';
isbvhodnvemrwvn 5y ago
Not fizzbuzz, but at one of my past companies we did a similar low complexity problem (use any language you fancy, google if you need to, you get a list of things, filter them by some attribute and group it by another, print out the list per group) during a phone screen, and people did fail it. Typically doesn't happen in later interviews, unless people have a complete mental breakdown due to stress.

boring_twenties 5y ago

Not FizzBuzz, and not me personally, but from someone I worked with and I have 0 reason to believe is lying: After a few disappointing responses to better questions, interviewer decides to give a "freebie" and asks for a function that determines whether an integer is odd or even. Candidate produces something that loops from 0 to the provided integer, incrementing two separate counters, then comparing them at the end. Interviewer manages to stammer out something like "are you sure there isn't a more efficient way to solve this?" Guy furrows his brow for a few minutes more, then a light bulb of sorts goes on, and he declares "Of course! ..." "... I only need one counter!"

Apocryphon 5y ago
That seems more like a failure of mathematical reasoning than in programming.
- kragen 5y ago
  I would be amazed to meet someone who, if asked whether 3042041 was odd or even, would follow this exponential-time procedure manually. Unless the candidate was such a person, it's a failure of programming.
  - Apocryphon 5y ago
    They just seem numerically illiterate to the point of not realizing how to determine odds or evens. Though perhaps, if they were more competent as a programmer, they would skip decimal math altogether and use the ol' bitwise operator trick.
    - boring_twenties 5y ago
      FWIW, gcc 10 generates identical code for `i % 2 != 0` and `i & 1 != 0` at -O1.
      - kragen 5y ago
        Yeah, but "doesn't know about the % operator" is a lot less damning than "can only think of an exponential-time algorithm to determine if a number is even". Like, this is terrible, but it's linear time: char s[32]; sprintf(s, "%d", n); char c = s[strlen(s) - 1]; return c == '0' || c == '2' || c == '4' || c == '6' || c == '8';

isbvhodnvemrwvn 5y ago

Not fizzbuzz, but at one of my past companies we did a similar low complexity problem (use any language you fancy, google if you need to, you get a list of things, filter them by some attribute and group it by another, print out the list per group) during a phone screen, and people did fail it. Typically doesn't happen in later interviews, unless people have a complete mental breakdown due to stress.