@sagar it will take O(n2) if all the elements of linked list are odd and
distinct......

On Sat, Jul 16, 2011 at 4:06 PM, sagar pareek <sagarpar...@gmail.com> wrote:

> 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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