Talk:TADM2E 3.28

I think this can be done in a single loop essentially taking two passes at once. Here is a Python snippet.

# O( n) - len( input_values)
length = len( input_values)		# the length of the input values
answers = [1]*length			# output values
frontIndex = 0				# the index at which we have calculated the product of all preceding indexes
frontProduct = 1			# the product of values preceding the front index
backProduct = 1				# the product of values following the back index
limit = length-1
backIndex = limit			# the index at which we have calculated the product of all following values
while( frontIndex < limit):
    frontProduct *= input_values[ frontIndex]
    backProduct *= input_values[ backIndex]
    frontIndex += 1
    answers[ frontIndex] *= frontProduct
    backIndex -= 1
    answers[ backIndex] *= backProduct

Both this suggestion and the solution given on the page still seem to be doing n-1 multiplications on n items, so they still seem to be n^2. --Murrayc2 (talk) 06:17, 8 November 2015 (EST)