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you mean list :: ch ?
When I wrote list :: ch1 :: ch2 I didn't compile then I thought It's wrong in scala. I don't have much experience in scala, It's almost a week I code in it
many compilers are optimized for tail-recursion which makes such functions less inefficient than one would normally expect them to be. For example, the tailRec keyword in Scala.
This is the solution that I eventually came up with and it still times out on test case 10. Interestingly, a simple for loop which prints out the characters of the strings also times out on that case and is about twice as long in execution on test case 9.
On my local machine, I get stack overflow way before the execution time problem.
Any suggestions?
Common source of long execution is the complexity of + operator (or append method) for strings. It's of Appending two strings, one of length N, other of M, is of complexity O(N+M). Here it's repeatitive addition, so it can be of quadratic complexity.
Sorry, I meant the onbe that you posted. Here is my version:
def Mingle(P:List[Char],Q:List[Char]):List[Char] = {
if(P.isEmpty)return List()
return P.head :: Q.head :: Mingle(P.tail,Q.tail)
}
def main(args: Array[String]) {
val P = readLine().toList
val Q = readLine().toList
println(Mingle(P,Q).mkString)
}
String Mingling
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you mean list :: ch ? When I wrote list :: ch1 :: ch2 I didn't compile then I thought It's wrong in scala. I don't have much experience in scala, It's almost a week I code in it
Something like
PS: Almost all the challenges are designed such that you don't need mutable variables to solve them.
Thank you it finally accepted. I'll check for differences those codes to understand what was the problem. Thank you again
Hello! I wonder why is this solution is better than mine:
It is tail recursive function, list concatenation has linear complexity
many compilers are optimized for tail-recursion which makes such functions less inefficient than one would normally expect them to be. For example, the tailRec keyword in Scala.
@serhii_nesteruk,
::has constant complexity, while++is linear.In your code you are applying
++n times. First on the list of length 1, then 2, then 3, and so on. So overall complexity will bethanks for the explanation!
This is the solution that I eventually came up with and it still times out on test case 10. Interestingly, a simple for loop which prints out the characters of the strings also times out on that case and is about twice as long in execution on test case 9. On my local machine, I get stack overflow way before the execution time problem. Any suggestions?
Hi @blowkj, sorry can't find your solution.
Common source of long execution is the complexity of
+operator (or append method) for strings. It's of Appending two strings, one of length N, other of M, is of complexityO(N+M). Here it's repeatitive addition, so it can be of quadratic complexity.Sorry, I meant the onbe that you posted. Here is my version: def Mingle(P:List[Char],Q:List[Char]):List[Char] = { if(P.isEmpty)return List() return P.head :: Q.head :: Mingle(P.tail,Q.tail) }
Note that I don't use the + operator. Thanks,
OK, got it now. Googling on tail recursion helped.