how is it possible in O(N) sir,wen array unordered??if u prove it u shall be
then next DONALD E KNUTH

On Thu, Aug 13, 2009 at 7:26 PM, Angad Karunan <[email protected]>wrote:

> why do you have to sort or do anything more if only one element repeats
> itself?
> just reading the array once shall give you the answer in O(n) time...
> so i guess there must be a solution better than O(n) time as the element
> repeats in a particular fashion.
> but cant think of anythin :P
>
>
> On Tue, Aug 11, 2009 at 11:51 PM, manish bhatia <[email protected]>wrote:
>
>> Well instead of using extra memory, we can in-place sort the arraay  in
>> O(nlog(n)) and then do an iteration (O(n)) to find out the repeated number.
>> But the catch is that number is repeated 2^i times. That is the hint we
>> should use
>>
>>  ------------------------------
>> *From:* ankur aggarwal <[email protected]>
>> *To:* [email protected]
>> *Sent:* Sunday, 9 August, 2009 5:47:32 PM
>> *Subject:* [algogeeks] Re: Finding repeated element in most efficient way
>>
>>
>> @richa..
>> ques is in complete i think .
>> there shud be some conditions given ..
>>
>> otherwise
>> hash them
>> but lots of space will b wasted..
>>
>>
>> or sort them
>>
>> try to put the conditions..
>>
>>
>>
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>>
>>
>
> >
>

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