Lazy Evaluation and Time Complexity

Question

I was looking around stackoverflow Non-Trivial Lazy Evaluation, which led me to Keegan McAllister\'s presentation: Why learn Haskell. In slide 8, he shows the minimum functi

Answer 1

Suppose minimum' :: (Ord a) => [a] -> (a, [a]) is a function that returns the smallest element in a list along with the list with that element removed. Clearly this can be done in O(n) time. If you then define sort as

sort :: (Ord a) => [a] -> [a]
sort xs = xmin:(sort xs')
    where
      (xmin, xs') = minimum' xs

then lazy evaluation means that in (head . sort) xs only the first element is ever computed. This element is, as you see, simply (the first element of) the pair minimum' xs, which is computed in O(n) time.

Of course, as delnan points out, the complexity depends on the implementation of sort.