yes , but that constraint is not provided by the interviewer , hence ,
solution of hash is not acceptable

On Fri, Aug 19, 2011 at 8:58 AM, Dheeraj Sharma <dheerajsharma1...@gmail.com
> wrote:

> hash map is the solution provided the elements lie in a predefined range..
>
> On Fri, Aug 19, 2011 at 8:46 AM, *$* <gopi.komand...@gmail.com> wrote:
>
>> true. I agree , we can use additional memory which will be constant
>> irrespective of counjt of elements.
>>
>> But using an hash wont be a constant memory as input can keep on varying.
>>
>> Thx,
>> --Gopi
>>
>>
>> On Fri, Aug 19, 2011 at 8:16 AM, Dipankar Patro <dip10c...@gmail.com>wrote:
>>
>>> O(1) space means constant space. It doesn't mean you can't use extra
>>> space.
>>> Refer here:
>>> http://stackoverflow.com/questions/2219109/what-does-this-mean-on-steps-and-o1-space
>>>
>>> According to the question you can definitely use a Hash Table for keeping
>>> hit record, as it will be a constant space (provided the range of numbers is
>>> known).
>>>
>>> In case the range of numbers is not known, BST will be close answer.
>>> Since only one element will be repeating, the process of making the BST can
>>> be stopped when the first repeating element is caught. BUT, this will be
>>> O(n) space, as the number of nodes in BST will be n-1 in worst case.
>>>
>>> On 19 August 2011 07:59, *$* <gopi.komand...@gmail.com> wrote:
>>>
>>>> only once
>>>>
>>>>
>>>> On Fri, Aug 19, 2011 at 7:57 AM, saurabh singh <saurab...@gmail.com>wrote:
>>>>
>>>>> The element is repeated only once or can be repeated k number of
>>>>> times??
>>>>>
>>>>> On Fri, Aug 19, 2011 at 7:50 AM, *$* <gopi.komand...@gmail.com> wrote:
>>>>>
>>>>>> I think we are using hash , which is like extra spaace , but as per
>>>>>> the question , O(s) = 1.
>>>>>>
>>>>>> Thx,
>>>>>> --Gopi
>>>>>>
>>>>>>
>>>>>> On Fri, Aug 19, 2011 at 2:15 AM, icy` <vipe...@gmail.com> wrote:
>>>>>>
>>>>>>> #!/usr/bin/ruby -w
>>>>>>> #array of unsorted positive integers
>>>>>>> # find the [only] one that is duplicated
>>>>>>>
>>>>>>> arr= [97,2,54,26,67,12,1,19,44,4,29,36,67,14,93,22,39,89]
>>>>>>> h = Hash.new(0)
>>>>>>>
>>>>>>> arr.each {|n|
>>>>>>>        h[n]+=1
>>>>>>>        (puts n; break) if h[n]==2
>>>>>>> }
>>>>>>>
>>>>>>> #output
>>>>>>> #67
>>>>>>>
>>>>>>> I hope this meets the requirements ;P
>>>>>>>
>>>>>>> On Aug 18, 3:15 pm, "*$*" <gopi.komand...@gmail.com> wrote:
>>>>>>> > How to find duplicate element (only one element is repeated) from
>>>>>>> an array
>>>>>>> > of unsorted positive integers..
>>>>>>> > time complexity .. O(n)
>>>>>>> > space .. o(1).
>>>>>>>
>>>>>>> --
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>>>>>>>
>>>>>>
>>>>>>
>>>>>> --
>>>>>> Thx,
>>>>>> --Gopi
>>>>>>
>>>>>>
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>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> Saurabh Singh
>>>>> B.Tech (Computer Science)
>>>>> MNNIT ALLAHABAD
>>>>>
>>>>>
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>>>>>
>>>>
>>>>
>>>>
>>>> --
>>>> Thx,
>>>> --Gopi
>>>>
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>>>
>>>
>>>
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>>
>>
>>
>> --
>> Thx,
>> --Gopi
>>
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>
>
>
> --
> *Dheeraj Sharma*
> Comp Engg.
> NIT Kurukshetra
> +91 8950264227
>
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-- 
Thx,
--Gopi

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