Citerus programming challenge
I'm very weak for programming challenges. Citerus held a contest where you could win an iPad if you would solve their problem. Since I'm a .NET developer, I'm not elegible for winning such a challenge but I love to meet a challenge just for the sake of it.
Given a string
And a collection of abbreviations
TDD, DDD, DI, DO, OO, UI, ANT, CV, IOC, LOC, SU, VO
What is the shortest string you could produce by removing abbreviations from it?
1) ...BIDDDOCDDD... (remove DDD) 2) ...BIDDDOC... (remove DDD again) 3) ...BIOC... (remove IOC)
This kind of problem is ideal for F#. The main problem is that you just can't remove abbreviations, because you won't find the shortest string. You will have to try to remove abbreviations in all orders to find the shortest string produced. I do this by recursive calls, where every abbreviation removal is a path in the tree. I solve the problem by taking the branch with the shortest result.
// In the string let s = "VOCDIITEIOCRUDOIANTOCSLOIOCVESTAIOCVOLIOCENTSU"
// Any occurrence of let abbreviations = ["TDD"; "DDD"; "DI"; "DO"; "OO"; "UI"; "ANT"; "CV"; "IOC"; "LOC"; "SU"; "VO"]
// Should be removed let rec remove (abbreviations : string List) (s : string) =
// Collect any version where one abbr is removed from s // Filter out those with no effect on s let collect = abbreviations |> List.map (fun abbr -> s.Replace(abbr, "")) |> List.filter (fun short_s -> short_s.Length < s.Length) // Select longest string of s1 and s2, or s1 if they're equal let min (s1 : string) (s2 : string) = if s1.Length <= s2.Length then s1 else s2 // Match result of abbreviations removal match collect with |  -> s | _ -> List.fold ( min ) s (collect |> List.map ( remove abbreviations ))
// Execute let solution = lazy ( s |> remove abbreviations ) Utilities.measureexecutiontime solution |> ignore
// Tests let tests = "HELLO" = ("HETDDLLDIO" |> remove abbreviations ) && // Two consecutive abbr "HELLO" = ("HEDTDDILLO" |> remove abbreviations ) && // Two nested abbr "HELLO" = ("HEDTDDILLO" |> remove abbreviations ) && // Removing TDD before DI "HELIUM" = ("HELANTDDDIUM" |> remove abbreviations ) && // Don't remove DI or TDD, remove ANT then DDD "" = ("DTDDIIOC" |> remove abbreviations ) && // Take TDD then IOC last DI will leave only empty string "B" = ("BIDDDOCDDD" |> remove abbreviations) // Some wierd example from the problem description
All in all, the algorithm code is 11 LOC, if you remove the comments, and runs for 320 ms on my machine.