i have solution with no extra space complexity but time complexity is >O(n)

traverse the list with a pointer ptr
if odd no encounter then traverse the remaining list with tmp pointer with
start point ptr->next and match the numbers with it....i hope it works :)

On Sat, Jul 16, 2011 at 10:10 AM, shady <sinv...@gmail.com> wrote:

> if hashing is allowed then it can be done in O(n)....... space complexity
> in this case again will be O(n)........ this won't work for large numbers...
>
>
> On Sat, Jul 16, 2011 at 1:58 AM, Nishant Mittal <
> mittal.nishan...@gmail.com> wrote:
>
>> @sagar... I know this solution but it was strictly asked to do in O(n)
>> time and O(1) space complexity and what if range of numbers is very
>> large....
>>
>>
>> On Sat, Jul 16, 2011 at 1:09 AM, sagar pareek <sagarpar...@gmail.com>wrote:
>>
>>> You just need to maintain the array for the odd words which encountered
>>> during traversing the list and using hashing this can be done :) but of
>>> course not in O(n) :(
>>>
>>> On Sat, Jul 16, 2011 at 12:51 AM, aseem garg <ase.as...@gmail.com>wrote:
>>>
>>>> Use a Hash Table.
>>>> Aseem
>>>>
>>>>
>>>>
>>>> On Sat, Jul 16, 2011 at 12:28 AM, shady <sinv...@gmail.com> wrote:
>>>>
>>>>> i don't think it is possible to do it in O(n)... rather not even in
>>>>> O(nlogn) without modifying the list
>>>>>
>>>>>
>>>>> On Fri, Jul 15, 2011 at 11:23 PM, Nishant Mittal <
>>>>> mittal.nishan...@gmail.com> wrote:
>>>>>
>>>>>> How will you delete duplicate odd numbers from a linked list in O(n)
>>>>>> time
>>>>>>
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>>> **Regards
>>> SAGAR PAREEK
>>> COMPUTER SCIENCE AND ENGINEERING
>>> NIT ALLAHABAD
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-- 
**Regards
SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD

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