ASRagab Here's a Scala implementation
arr.flatMap(List.fill(num)(_))
christopher_lil1 Similar to Haskell: concatMap (\num -> replicate n num) arr
MichaelQQ also works: concatMap (replicate n) arr
eloyekunle Also works:
f n = concatMap (replicate n)
idan5x Also works:
f = concatMap . replicate
josemanzoli I think this way is much more readable, even for people that are used to see functional programming codes!
arr.flatMap(e => List.fill(num)(e))
avvin josemanzoli how to excel in scala any help or tips ?
byron_lema [deleted]
baldockja Could someone explain why we have to pass the parameters as "(a)(b)" instead of "(a, b)".
If this is because "List.fill(a)" returns a function which is then passed "b", then why in the documentation is this code fragment refered to as if it is function in and of itself (http://www.scala-lang.org/api/current/scala/collection/generic/SeqFactory.html#fillA(elem:=>A):CC[A])?
Thanks.
mayonesa It is currying.
fill(a)orfill(a)(b)are both functions. The only difference is thatfill(a)(b)returns aListwhilefill(a)returns ab => Listfunction (e.g., a function that when givenbreturns aList).
_s_batra Haskell:-
f n arr = do [num | num <- arr , a <- [1..n]]
dillonheins Hi there. Could you possibly explain what is happening here? Is
a <- [1..n]a predicate or just a normal list? What I do understand is that num is the output, you are mapping arr to num but then how does mapping[1..n]toaaffect the output?jonathandygert When you do list comprehensions in Haskell and draw from multiple lists (by doing
x <- list), then the result is a list doing the thing in front for all posible combinations from the lists. Here's an example:[(a, b) | a <- [1, 2, 3], b <- [4, 5]]
gives you
[(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)]
In _s_batra's example, the
ais kinda unused (which also means that by convention it should be named_), but it simply repeats thenumfor each element in[1..n](akantimes).WanpengWang Cartesian product
eliaszx that is also analogous to nested loops in imperative languages(C#, Java) something like:
foreach(var num in arr) { for(var a=1; a <=n; a++) { yield return num; } }
chesschingis Much more beautiful/mathematical!
fairjm If you use
fsharp.Code followed is for input.let s = Console.ReadLine() |> int let list = [ let mutable key = Console.ReadLine() while not (key = null) do yield key key <- Console.ReadLine() ]mwojtkowski Thanks !!!
stt106 Nice!!!
edward_trenchar1 spot on - thanks!
waahhoo Here's a more functional way to read the inputs:
let n = Console.ReadLine() |> int Seq.initInfinite (fun _ -> Console.ReadLine()) |> Seq.takeWhile (String.IsNullOrWhiteSpace >> not) |> Seq.toList
arche3r Thanks!
vaskarvelas Nice!!! :-)
lizzael_v_c Thanks!
borsh Thanks a lot!
vaskarvelas [deleted]
FlavioFernandes4 Another scala impl:
for{e <- arr; _ <- 1 to num} yield e
mayonesa which is probably syntactic sugar for
arr.flatMap(e => (1 to num).map(_ => e))
wamaral Haskell is so good for these kind of problems
f = concatMap . replicate
chesschingis Ikr!
adrielradicchi Here's Elixir implementation
defmodule Solution do def times(n, element) do if (n > 1) do IO.puts element times(n-1, element) else IO.puts element end end def elements(array) do [_ | rest ] = array Enum.map(rest, fn(x) -> times(List.first(array), x) end) end def main() do IO.read(:stdio, :all) |> String.split |> Enum.map(&String.to_integer(&1)) |> elements end end Solution.main
darren_m_k My version in F#:
(open, readline parts are excluded as they are already shown in other comments)
List.iter (printfn "%s") (List.collect (fun x -> List.replicate num x) list)
c_song Struggled a lot for input handling, here is some reference for those having a hard time with erlang.
input(L) -> case io:fread("", "~d") of {ok, [N]} -> input([N|L]); _ -> lists:reverse(L) end. output(L) -> [io:fwrite("~w~n", [H]) || H<-L].
The actual solution with list comprehension is easy as haskell:
[N || N <- X , _ <- lists:seq(1,S)]
giusepanzuto thank you
ocramius The exercise is wrong/mis-specified.
Mentions "The relative positions of the values should be the same as the original list provided in the input.", but the test cases seem to expect a sorted list.
Discovered it by accident by looking at existing solutions after implementing my own (working against all my own test scenarios) :-(
reggaeguitar This should definitely be specified in the problem description, I'm really glad I read this comment before tearing my hair out.
VardiN I am new to functional programming, But still wanted to try these problems. Can some one tell me what is the logic To break the read cycle in RACKET
I mean in sample test case, after taking 4 how should I break it
abhiranjan Check this :)
(define (read-list) (let ([x (read)]) (if (eof-object? x) (list) (cons x (read-list)))))VardiN thanks :)
kah5368 I've spent countless hours trying to figure this out... thanks a million!
harit66 How did you add code so neatly? mine is screwed up
abhiranjan We follow markdown format. Indenting the code segment by 4 whitespaces (1 tabspace) will work.
zengmor Thank you!
atharvashukla Thank you! Had the same problem.
johanirvall Very nice!
giaym im amazed i managed to compelte it %("$&( odd languages
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