Make membership linear-time overall

Problem

Implement all_present(xs, ys), returning whether every distinct value in xs occurs in ys. Construct at most one auxiliary collection so expected time is O(len(xs) + len(ys)).

Starter code

def all_present(xs, ys):
    pass

Reveal answer or reference solution

def all_present(xs, ys):
    lookup = set(ys)
    return all(x in lookup for x in xs)

Public tests

all_present([1,2,2], [3,2,1]) → True
all_present([1,4], [1,2,3]) → False
all_present([], []) → True

Local history

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Python Practice