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Concise algorithm solution in Speedy category for Frequency Sorting by Igor_Sekretarev
from collections import Counter
from typing import List
def frequency_sorting(numbers: List[int]) -> List[int]:
count = Counter(numbers)
numbers.sort(key=lambda x: (-count[x], x))
return numbers
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"
print("Coding complete? Click 'Check' to earn cool rewards!")
May 6, 2021
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