Sorting: by frequency

sorted can sort by anything you want via a key function. Return a tuple and Python sorts by the first element, then the second as a tiebreaker:

from collections import Counter
counts = Counter([1, 1, 2, 2, 2, 3])   # {1: 2, 2: 3, 3: 1}

sorted([3, 1, 2], key=lambda x: (-x, x))  # negate to sort DESCENDING

A Counter tallies how often each value appears. Negating a number flips the sort direction, so key=(-count, value) means most frequent first, ties broken by the smaller value.

Your task

Write frequency_sort(nums) that returns the same numbers, sorted by frequency descending, with ties broken by the number ascending.

frequency_sort([1, 1, 2, 2, 2, 3])   # [2, 2, 2, 1, 1, 3]
frequency_sort([4, 4, 1])            # [4, 4, 1]
frequency_sort([])                   # []

Hint: build counts = Counter(nums), then sorted(nums, key=lambda x: (-counts[x], x)).