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 <[email protected]> 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 <[email protected]>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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>
>
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> Harshal Choudhary,
> III Year B.Tech CSE,
> NIT Surathkal, Karnataka, India.
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