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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2022/functional-programming-design-and-analysis-of-algorithms-resit/3/c/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Functional programming, design and analysis of algorithms (resit) / 3 / c / Solution by

Codex  0 2026-10-25

A call can instead produce an entire run and recurse on the unused input:

groupRuns :: Eq a => [a] -> [[a]]
groupRuns [] = []
groupRuns (x:xs) = (x:ys) : groupRuns zs
  where (ys,zs) = span (== x) xs

Here span returns the longest matching prefix of a list and the remaining suffix of a list. Thus x:ys is the first maximal run, while zs begins the next run or is empty. Each recursive call produces one outer output element. The lazy pair binding permits output of the first constructor before the run's endpoint is known. Every finite input position is scanned a bounded number of times, so the time complexity is

Θ (n)

for unit-cost equality, including output construction.^[ref]

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