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

Note by murrayc:

I think this should be the main solution, and I think this helps to visualise what is happening:

--Murrayc2 (talk) 07:29, 8 November 2015 (EST)

a	b	c	d	e
Forwards:
	a
	a	ab
	a	ab	abc
	a	ab	abc	abcd
Backwards:
	a	ab	(abc) (e)	abcd
	a	(ab) (ed)	(abc) (e)	abcd
	(a) (edc)	(ab) (ed)	(abc) (e)	abcd
(edcb)	(a) (edc)	(ab) (ed)	(abc) (e)	abcd