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Comment by Thrymr | Hacker News Reader

  sum([x for x in xrange(1, 101) if x % 3 == 0 and x % 5 == 0])

edit: actually sum the list. Of course, it's even shorter if you realize you can write it as:

  sum([x for x in xrange(1, 101) if x % 15 == 0])

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You're better off making that a generator:

sum((x for x in xrange(1, 101) if x % 3 == 0 and x % 5 == 0))

Notice the parenthesis instead of square brackets. Your version actually creates a list in memory. A generator generates the items one by one and doesn't need to store them all at the same time.

You can just write it as

  sum(x for x in xrange(1,101) if x % 3 == 0 and x % 5 == 0)

and skip the inner set of parenthesis; the parenthesis on a generator expression are not needed if it is the only argument to a function.

Can't you all think for a bit? All of These solutions are O (n) in both time and space... there is solution that will take O (1) time and space - and I think that you learned it in math class at (latest) high school (hint: arithmetic progression)

Seriously? The discussion is about language features, not optimal algorithms.

One could imagine the question was actually "sum BusyBeaver(n) for n divisible by 3 and 5 between 1 and 100", i.e. no closed form. In Python:

  sum(BusyBeaver(x) for x in xrange(1,101) if x % 3 == 0 and x % 5 == 0)

(And yes, if one was being really pedantic, one could replace the condition with x % 15 == 0.)

The thing is that people start using those libraries w/o thinking before - and theirs' solutions end up doing much more work than is needed

This could be said of most anything high level.

true, but xrange(1,101) is so tiny, who cares!

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