How to solve longEnough?. In Haskell -

longenough n xs: checks if list has more n elements.

examples:

• longenough 2 [1..5] == true
• longenough 3 [1,2,3] == false
• longenough 0 [] == false
• longenough 20 [1..] == true

i guess homework , still learning basics i'll start giving hints using recursion instead of foldr (as @dfeuer proposed):

first start noting obvious cases:

• if xs = [] result false (assuming don't care negative n in strange ways)
• if n = 0 , xs non-empty it's true
• in other cases have
  • n > 0
  • xs has more 1 element

maybe have recursive idea break last case down?

here skeleton:

longenough :: int -> [a] -> bool longenough _ []     = false longenough 0 _      = true longenough n (_:xs) = let n' = (n-1) in undefined 

for cases - if look closely you'll see added more hints on solution.

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ps

• maybe want think negative n , should happen ... did not here
• if know foldr should try implement using foldr too

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solution

seems there no more feedback coming op guess can post solution start with:

longenough :: int -> [a] -> bool longenough _ []     = false longenough 0 _      = true longenough n (_:xs) = longenough (n-1) xs 

(not left do...)

here mentioned test-cases:

λ> longenough 2 [1..5] true λ> longenough 3 [1,2,3] false λ> longenough 0 [] false λ> longenough 20 [1..] true