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First solution in Clear category for Min and Max by aureooms
def min ( first , *args , key = lambda x : x ) :
if not args :
args = iter( first )
if not args : raise ValueError( "min() arg is an empty sequence" )
first = next( args )
smallest = first
for candidate in args :
if key( candidate ) < key( smallest ) :
smallest = candidate
return smallest
def max ( first , *args , key = lambda x : x ) :
if not args :
args = iter( first )
if not args : raise ValueError( "max() arg is an empty sequence" )
first = next( args )
largest = first
for candidate in args :
if key( candidate ) > key( largest ) :
largest = candidate
return largest
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"
March 27, 2015
