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Second solution in Clear category for Min and Max by aminami1127
def min(*args, **kwargs):
key = kwargs.get("key", lambda x: x)
values = iter(args[0]) if len(args) == 1 else iter(args)
min_value = next(values)
for i in values:
if key(i) < key(min_value):
min_value = i
return min_value
def max(*args, **kwargs):
key = kwargs.get("key", lambda x: x)
values = iter(args[0]) if len(args) == 1 else iter(args)
max_value = next(values)
for i in values:
if key(i) > key(max_value):
max_value = i
return max_value
if __name__ == '__main__':
#These "asserts" using only for self-checking and not necessary for auto-testing
assert max(3, 2) == 3, "Simple case max"
assert min(3, 2) == 2, "Simple case min"
assert max([1, 2, 0, 3, 4]) == 4, "From a list"
assert min("hello") == "e", "From string"
assert max(2.2, 5.6, 5.9, key=int) == 5.6, "Two maximal items"
assert min([[1, 2], [3, 4], [9, 0]], key=lambda x: x[1]) == [9, 0], "lambda key"
July 6, 2015
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