Frequency map with deterministic ties

Problem

Implement most_frequent(xs). Return the value with highest frequency; break ties by returning the smallest value. Return None for an empty input.

Starter code

def most_frequent(xs):
    pass

Reveal answer or reference solution

def most_frequent(xs):
    if not xs:
        return None
    counts = {}
    for x in xs:
        counts[x] = counts.get(x, 0) + 1
    return min(counts, key=lambda x: (-counts[x], x))

Public tests

most_frequent([3, 1, 3, 1, 2]) → 1
most_frequent([]) → None

Local history

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Data Structures Practice