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sorted dictionary solution in Clear category for Nearest Value by marsal25
def nearest_value(values: set, one: int) -> int:
d={}
for i in sorted(values,reverse=True):
if i=one:
d[i-one]=i
return d.get(min(d))
if __name__ == '__main__':
print("Example:")
print(nearest_value({4, 7, 10, 11, 12, 17}, 9))
# These "asserts" are used for self-checking and not for an auto-testing
assert nearest_value({4, 7, 10, 11, 12, 17}, 9) == 10
assert nearest_value({4, 7, 10, 11, 12, 17}, 8) == 7
assert nearest_value({4, 8, 10, 11, 12, 17}, 9) == 8
assert nearest_value({4, 9, 10, 11, 12, 17}, 9) == 9
assert nearest_value({4, 7, 10, 11, 12, 17}, 0) == 4
assert nearest_value({4, 7, 10, 11, 12, 17}, 100) == 17
assert nearest_value({5, 10, 8, 12, 89, 100}, 7) == 8
assert nearest_value({-1, 2, 3}, 0) == -1
print("Coding complete? Click 'Check' to earn cool rewards!")
Nov. 13, 2020
