Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Functional programming, design and analysis of algorithms / 4 / b / Solution

Flatten the days, emit both participants of every match and remove repeated names:

teams :: [[Match]] -> [Team]
teams days = unique [t | Match a _ b _ <- concat days, t <- [a,b]]
  where
    unique []     = []
    unique (t:ts) = t : unique (filter (/= t) ts)

Every emitted name belongs to some match, and every participating team enters the intermediate list. At each recursive step, unique emits its first remaining name once and deletes precisely its later occurrences. Induction therefore proves both coverage and absence of duplicates. The output preserves first appearance; no particular team ordering was required. This simple list implementation can take quadratic time in the number of participant occurrences, which is permissible here; the question imposes no efficiency target.