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First solution in Clear category for Min and Max by mistymagich
def min(*args, **kwargs):
return __common('min', args, kwargs)
def max(*args, **kwargs):
return __common('max', args, kwargs)
def __common(minmax, args, kwargs):
key = kwargs.get("key", None)
if len(args) == 1:
args = args[0]
args = list(args)
index = 0
m = args[0] if key == None else key(args[0])
for i in args:
_index = args.index(i)
c = i if key == None else key(i)
if (minmax == 'max'):
if (m < c):
m = c
index = _index
elif (minmax == 'min'):
if (m > c):
m = c
index = _index
return args[index]
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"
Oct. 16, 2015