Difference between revisions of "Talk:TADM2E 3.28"

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(This actually seems to be doing 4 multiplications n times, so it is O(n).)
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     answers[ backIndex] *= backProduct
 
     answers[ backIndex] *= backProduct
 
  </source>
 
  </source>
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Note by murrayc:
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I think this should be the main solution, and I think this helps to visualise what is happening:
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{| class="wikitable" |
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! a
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! b
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! c
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! d
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! e
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|-
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| Forwards:
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|-
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|
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| '''a'''
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|
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|
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|
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|-
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|
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| a
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| '''ab'''
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|
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|
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|-
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|
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| a
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| ab
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| '''abc'''
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|
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|-
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|
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| a
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| ab
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| abc
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| '''abcd'''
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|-
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| Backwards:
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|-
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|
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| a
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| ab
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| (abc) '''(e)'''
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| abcd
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|-
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|
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| a
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| (ab) '''(ed)'''
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| (abc) (e)
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| abcd
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|-
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|
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| (a) '''(edc)'''
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| (ab) (ed)
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| (abc) (e)
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| abcd
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|-
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| '''(edcb)'''
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| (a) (edc)
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| (ab) (ed)
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| (abc) (e)
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| abcd
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|-|}
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--[[User:Murrayc2|Murrayc2]] ([[User talk:Murrayc2|talk]]) 07:29, 8 November 2015 (EST)

Revision as of 12:29, 8 November 2015

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
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