For example: ʺaaaabbaabʺ->[(‘a’,4),(‘b’,2),(‘a’,2),(‘b’,1)] Its need to be done using FOLDR through one pass of the list, without using ( ).

Here what I have so far

task2 (x:xs) = foldr (\c [(symbol, count)] -> if symbol == c then [(symbol, count 1)] else [(symbol, count)]) [(x, 1)] xs

The problem is I don't really understand how to make it go to the next element of the list after 'if' statement is False

CodePudding user response：

Writing the step function as an inline lambda expression is probably not the best possible move. It can be made to work, but that leads to a very long line of code.

It is easier to write the step function separately, like this:

task2 :: String -> [(Char,Int)]
task2 cs = foldr stepFn [] cs
  where
    stepFn c      []          =  [(c,1)]  -- simple case
    stepFn c ((c1,n1) : ps)   =           -- please try to write the rest ...

if (c == c1) then (c1,1 n1) : ps else (c,1) : (c1,n1) : ps

Testing:

$ ghci
 GHCi, version 8.8.4: https://www.haskell.org/ghc/  :? for help
 λ> 
 λ> :load q69871708.hs
 [1 of 1] Compiling Main             ( q69871708.hs, interpreted )
 Ok, one module loaded.
 λ> 
 λ> task2 "aaaabbaabrrrzz"
 [('a',4),('b',2),('a',2),('b',1),('r',3),('z',2)]
 λ> 
 λ> task2 "a"
 [('a',1)]
 λ> 
 λ> task2 ""
 []
 λ>