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sorted solution in Clear category for Frequency Sorting by alek-pol
from typing import List
def frequency_sorting(numbers: List[int]) -> List[int]:
return sorted(numbers, key=lambda x: (-numbers.count(x), x))
if __name__ == "__main__":
print("Example:")
#print(frequency_sorting([1, 2, 3, 4, 5]))
# These "asserts" using only for self-checking and not necessary for auto-testing
assert frequency_sorting([1, 2, 3, 4, 5]) == [1, 2, 3, 4, 5], "Already sorted"
assert frequency_sorting([3, 4, 11, 13, 11, 4, 4, 7, 3]) == [
4,
4,
4,
3,
3,
11,
11,
7,
13,
], "Not sorted"
assert frequency_sorting([99, 99, 55, 55, 21, 21, 10, 10]) == [
10,
10,
21,
21,
55,
55,
99,
99,
], "Reversed"
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Aug. 25, 2021
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