Next coding challenge..

[Ben Hutchison]

This one is based on something Eric Torreborre tweeted a few years
back. It will sharpen your Scala Sollections Fu:

You have a finite Seq[Int], unordered and may contain duplicates.
Write a function, Seq[Int] => Seq[(Int, Int)], that converts into an
interval representation, where each contiguous series of integers is
represented by an interval tuple (start, end), and the tuples are
ordered in ascending order.

For example:

In: Seq(4, 1, 4, 2, 5, 8, 2, 6)

Out: Seq((1, 2), (4, 6), (8, 8))

This can be done in a few of lines of Scala, so perhaps wait on
posting solutions until friday?

-Ben

[etorreborre]

That's funny because I indeed remember the problem which was "compressing" a long list of ids into just a few ranges but I don't remember how I did it :-).

Eric.

[Ken Scambler]

The collections classes have sprouted a lot of nifty functionality since then too, so the solution might be even more concise now!

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[Jem Mawson]

Let me rephrase the challenge:

def ranges(ints: Seq[Int]): Seq[(Int, Int)]

assert(ranges(Seq()) == Seq()) // empty

assert(ranges(Seq(1)) == Seq((1,1))) // single value

assert(ranges(Seq(1,2,3)) == Seq((1,3))) // single range

assert(ranges(Seq(1,2,3,9)) == Seq((1,3),(9,9))) // 1 range, 1 value

assert(ranges(Seq(1,2,3,7,8,9)) == Seq((1,3),(7,9))) // two ranges

assert(ranges(Seq(5,7,3)) == Seq((3,3),(5,5),(7,7))) // distinct values unordered

assert(ranges(Seq(6,4,5)) == Seq((4,6))) // range unordered

I look forward to seeing some interesting answers. Especially a succinct one from Ben.

Cheers

Jem