Talk:99 questions/11 to 20
The prototype for repli in problem 15 is
repli :: [a] -> Int -> [a]
Because the second parameter is the number of times to replicate, it discourages the use function composition. I mean that if you swapped the parameters you could write it pointfree:
repli :: Int -> [a] -> [a] repli n = foldr (\ x xs -> replicate n x ++ xs) 
This would also match the way replicate is defined:
replicate :: Int -> a -> [a]
So, I suggest modifying problem 15 by swapping the parameters to repli in the example and the solution.
I made an edit to this page. I removed the following solution to problem 18:
slice xs i j = [xs!!(i-1)..xs!!(j-1)]
slice [1,3,6,3,1,6,7,8,3,2,4,76,8] 4 5 ==  Thanks to pixel for pointing this out.
The solution to problem 20 seems to be using 0-based indexing, whereas the question called for 1-based indexing in the other languages. This can be easily fixed:
removeAt :: Int -> [a] -> (a, [a]) removeAt k l = (elementAt l k, take (k-1) l ++ drop k l)
using elementAt from a previous problem.
or if you want to express that 1-based indexing is silly,
removeAt n+1 xs = (xs!!n,take n xs ++ drop (n+1) xs)