Hi

I think multi hash table will solve this problem.

Thanks
Supraja J

On Thu, Jun 2, 2011 at 1:37 PM, Harshal <hc4...@gmail.com> wrote:

> But I think the solution required is O(n) (one pass). So, O(nlogn) is not
> what the author wants anyway. Hashing is an option but we dont know the
> range of the elements in the array. So, in case when all keys hash into the
> same place, the worst case is still O(n^2).
> suppose our hash function is h(n) = n mod 10.
> array: 2,12,32,52,42.
>
>
>
> On Fri, Jun 3, 2011 at 1:01 AM, Harshal <hc4...@gmail.com> wrote:
>
>> you can check if the element to be inserted next is same as the node value
>> while inserting itself. Why to do a separate inorder traversal..
>>
>>
>> On Fri, Jun 3, 2011 at 12:56 AM, Supraja Jayakumar <
>> suprajasank...@gmail.com> wrote:
>>
>>> Hi
>>>
>>> Inserting into BST and break on same element will detect duplicates.
>>> But in any case, this is still O(2n) ? - once for inserting them in to
>>> the tree
>>> and another for the inorder read.
>>>
>>> Correct me if wrong.
>>>
>>> Rgds
>>> Supraja J
>>>
>>>
>>> On Thu, Jun 2, 2011 at 1:11 PM, Harshal <hc4...@gmail.com> wrote:
>>>
>>>> @Anurag: XOR wont work here, 1 element is repeated, not 1 element is
>>>> unique. Read the question again.
>>>>
>>>> Keep inserting elements in a BST and break once you find the same
>>>> element. O(nlogn)
>>>>
>>>>
>>>> On Fri, Jun 3, 2011 at 12:38 AM, Anurag Narain 
>>>> <anuragnar...@gmail.com>wrote:
>>>>
>>>>> take X-OR of all the elements.....the one which has no duplicate will
>>>>> be left and rest all will be reduced to zero.
>>>>>
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>>>>
>>>>
>>>>
>>>> --
>>>> Harshal Choudhary,
>>>> III Year B.Tech CSE,
>>>> NIT Surathkal, Karnataka, India.
>>>>
>>>>
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>>>
>>>
>>>
>>> --
>>> U
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>>
>>
>>
>> --
>> Harshal Choudhary,
>> III Year B.Tech CSE,
>> NIT Surathkal, Karnataka, India.
>>
>>
>>
>
>
> --
> Harshal Choudhary,
> III Year B.Tech CSE,
> NIT Surathkal, Karnataka, India.
>
>
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U

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