Difference between revisions of "Talk:TADM2E 3.28"
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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" | | ||
| + | ! a | ||
| + | ! b | ||
| + | ! c | ||
| + | ! d | ||
| + | ! e | ||
| + | |- | ||
| + | |colspan="5"| Forwards: | ||
| + | |- | ||
| + | | | ||
| + | | '''a''' | ||
| + | | | ||
| + | | | ||
| + | | | ||
| + | |- | ||
| + | | | ||
| + | | a | ||
| + | | '''ab''' | ||
| + | | | ||
| + | | | ||
| + | |- | ||
| + | | | ||
| + | | a | ||
| + | | ab | ||
| + | | '''abc''' | ||
| + | | | ||
| + | |- | ||
| + | | | ||
| + | | a | ||
| + | | ab | ||
| + | | abc | ||
| + | | '''abcd''' | ||
| + | |- | ||
| + | |colspan="5"| 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 | ||
| + | |-|} | ||
| + | |||
| + | --[[User:Murrayc2|Murrayc2]] ([[User talk:Murrayc2|talk]]) 07:29, 8 November 2015 (EST) | ||
Latest revision as of 12:30, 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:
| 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 |