That's just really sad, that there are people that would be interviewing for a coding position and it turns out they can't even do that in under a minute or so.
How do they expect to do their jobs ?
How do they expect to do their jobs ?
First thing I'd ask is where is the camera :)
I can see how for a junior coding position this might be an appropriate question, say people fresh out of school, and then 1 to 10 minutes might be acceptable.
But 4 modulo statements (or 3, if you think about the problem a bit longer) and a loop ?
10 minutes ?
That's pretty slow. I understand there is a lot more to programming than coding up a simple solution like this but as problems come it is really a very simple one and if someone would take 10 minutes to put this together I'd be a bit worried about throughput, and probably about experience as well.
Now if they are actually struggling for 10 minutes, that's another matter.
Now suppose the modulo test is really expensive. Suppose we're doing something complicated with large records on disk and we want to do this is that's true, something else if the other is true, etc, etc, just as in FizzBuzz.
How would you restructure your code so it's still obvious to a maintainer what it's doing, but so that it avoids doing more modulo operations than necessary.
You see, all these trivial exercises can be used as starting points for deeper conversations about aspects of coding.
main()
{
int i;
int ncounters = 2;
int counters[2] ;
int presets[2] = { 5, 3};
int n; // number of counters that tripped
int j;
char * strings[2] = { "fizz", "buzz" };
// first time, copy presets to counters
for (j=0;j<ncounters;j++) {
counters[j] = presets[j];
}
for (i=1;i<=100;i++) {
n = 0; // reset number of counters that have zeroed
for (j=0;j<ncounters;j++) {
counters[j] = counters[j] - 1;
// output relevant string when counter trips
if (counters[j] == 0) {
// separate strings by dashes if more than one counter trips
if (n != 0) {
printf("-");
}
n++;
counters[j] = presets[j];
printf("%s",strings[j]);
}
}
// no counter tripped, just output the number
if (n == 0) {
printf("%d",i);
}
printf("\n");
}
}
No modulo operations.
By asking the question that way you'd get solutions that you're not really looking for.
int c3 = 1;
int c5 = 1;
for ( int i = 1; i <= 100; i++, c3++, c5++ ){
if ( c3 == 3 ){ printf( "FIZZ" ); c3 = 0; }
if ( c5 == 5 ){ printf( "BUZZ" ); c5 = 0; }
if ( c3 && c5 ){ printf( "%d", i ); }
}Elegant solution, but fails to meet the problem specification.
I agree it's a good starting point, but surely it's still just getting rid of the ridiculously bad programmers rather than anything else.