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Comment by azylman | Hacker News Reader

Solution in CoffeeScript:

    add1 = (arr, val, n) ->
      n = arr.length if n is 0
      arr = arr.reverse() if n < 0
      count = 0
      for num, i in arr
        break if count > (Math.abs n)
        arr[i]++ if num is val
        count++
      return if n < 0 then arr.reverse() else arr

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another double reverse... I guess coffeescript really does want to be ruby :)

There was only two ways I could think of to do this without the double reverse.

One would introduce more complex branches (looping backwards in CoffeeScript ain't easy) and the other would be using Underscore.js's range function to do the same thing - that solution would be basically identical to this, except without the reverses, only I'd consider that cheating.

Here's my cheating version that uses a third-party library:

    add1_cheating = (arr, val, n) ->
      n = arr.length if n is 0
      start = if n < 0 then arr.length else 0
      end = if n < 0 then 0 else arr.length
      step = n/(Math.abs n)
      n = Math.abs n
      count = 0
      for i in _.range start, end, step
        break if count > n
        arr[i]++ if arr[i] is val
        count++
      return arr

People are just lazy since it usually doesn't matter.

array[i]++ for i in [-1..-array.length]

Here's my version without using reverse or underscore. If coffeescript supported reverse loops or negative indices, this would be a lot cleaner.

    add1 = (arr, val, n)->
      d = Math.abs(n||1)/(n||1)
      n = if n is 0 then arr.length else Math.abs n
      i = if d < 0 then arr.length - 1 else 0
      end = if d < 0 then 0 else arr.length

      while n and i isnt end
        num = arr[i]
        if num is val
          n--
          arr[i]++
        i = i + d
      arr

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