I assume the no.s are in the range 1 to n.
Now XOR all the elements together.
Now XOR the result with all numbers from 1 to n in a loop.

The Repeated no. occurs thrice(the result is Repeated number) and all others
get XORed twice(give 0 result).

Thus the result will be the repeated number .

*Regards

Sanju

Happy to Help :)*



On Thu, Aug 18, 2011 at 8:42 PM, *$* <[email protected]> wrote:

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