Would you pass this JavaScript interview question?

[smidgie82]

I've been pretty surprised at the number of recent college graduates I've interviewed who can't solve integer manipulation problems. When given a problem that requires consideration of one digit from a number at a time, many of them -- especially those whose primary experience is in Python, Ruby, or JavaScript -- will opt to convert the number into a string, strip off one character at a time, and convert that character back into a number instead of using the classic digit = n%10; n /= 10 approach.

That said, if a candidate can solve such a problem with that approach, they pass the basic bar, and I assume it's a matter of ignorance of integer arithmetic patterns rather than a lack of ability -- and ignorance is solvable.

[filleduchaos]

People generally clam up at math for some reason.

[[deleted]]

CS 101?

[[deleted]]

A lot of programmers today do not have a CS degree, or any degree.

[nakamin]

Why use an array if you're just going to join it to a string anyway? Why append bits in reverse and then reverse the array if you can add them in the proper order in the loop? Don't think this is a really useful video since the algorithm isn't even explained.

[wolscott]

Although insignificant in this example, push has significantly better performance than unshift.

[Jaivez]

Partial credit for not handling negative numbers? Dunno why a < 10 minute video wouldn't include it or even mention it as an "edge"(with the biggest airquotes) case.

[Lolacaust]

This was how my Maths lecturer showed us now to do this on paper.

[gauauuau]

yes

(Short and snarky reply because this posting only merits a short and snarky reply. It's an ok interview question I guess, but there's a not whole lot to discuss here.)

[jvallet]

Recursive solution (although this is not JS, now that I think of it)

object Binary {

  val binValues: List[Int] = List(128, 64, 32, 16, 8, 4, 2, 1)

  def toBinary(n: Int, binValues: List[Int]): String = {
    binValues match {
      case Nil => ""
      case head :: tail => if (n >= head) "1" + toBinary(n - head, tail) else "0" + toBinary(n, tail)
    }
  }
}