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First solution in Clear category for Min and Max by jszymk
def count_iterable(i):
return sum(1 for e in i)
def min(*args, **kwargs):
key = kwargs.get("key", None)
if count_iterable(args) > 1 :
ret = args[0]
for i in args :
if key != None :
if key(i) < key(ret) :
ret = i
else :
if i < ret :
ret = i
else :
arg = list(args[0])
ret = arg[0]
for i in arg :
if key != None :
if key(i) < key(ret) :
ret = i
else :
if i < ret :
ret = i
return ret
def max(*args, **kwargs):
key = kwargs.get("key", None)
if count_iterable(args) > 1 :
ret = args[0]
for i in args :
if key != None :
if key(i) > key(ret) :
ret = i
else :
if i > ret :
ret = i
else :
arg = list(args[0])
ret = arg[0]
for i in arg :
if key != None :
if key(i) > key(ret) :
ret = i
else :
if i > ret :
ret = i
return ret
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. 17, 2016
