One more approach using XOR to find the element repeated thrice.
Complexity: O(n).
Space :0

http://codepad.org/p82TGhjR

main()
{
  int arr[]= {5,3,3,1,5,5,7,7,8,8};
  int len, set_bit_no, x,y,i;
  int xor, prev;
  len = sizeof(arr)/sizeof(arr[0]);
  xor = arr[0];
  x = y=0;

  for(i=1;i<len;i++)
  {
    xor ^= arr[i];
  }

  printf("xor:%d\n",xor);

  for(i=0;i<len;i++)
  {
    xor ^= arr[i];
    if(xor == 0 && prev == arr[i])
    {
       printf("Found:%d\n", arr[i]);
       break;
    }
    prev = arr[i];
  }
}



On Thu, Jul 8, 2010 at 6:46 AM, jalaj jaiswal <jalaj.jaiswa...@gmail.com>wrote:

> @ any solution less then nlogn would do + O(1) space
>
>
> On Thu, Jul 8, 2010 at 12:38 AM, souravsain <souravs...@gmail.com> wrote:
>
>> @jalaj
>>
>> Are we looking for a better than )(nlogn) time and O(1) space
>> solution? What if our target?
>>
>> If a solution is required simple, then as mentioned by Satya, sort the
>> numbers in O(nlogn) time and scan once in O(n) time. So we get the
>> number repeated 3 times in O(nlogn) time and O(1) space.
>>
>> Sourav
>>
>> On Jul 7, 7:36 pm, Priyanka Chatterjee <dona.1...@gmail.com> wrote:
>> > > I am sceptical whether any XOR solution  exits for your question. But
>> if
>> > > the question is modified as :
>> >
>> > > *Only one number repeats once,* some no.s repeat twice and only one
>> number
>> > > repeat thrice, here is the XOR solution for that.
>> >
>> > > suppose the sample array is A[]={1, 3,3,5,5,5, 7,7,8,8}
>> > > in the example 1 repeats once and 5 repeats thrice.
>> >
>> > > 1>let T= XOR( all elements)= 1^5. (all elements occurring even no of
>> times
>> > > nullify)   -O(N)
>> >
>> > > (  let x=1, y=5
>> > > As we know the no. repeating once and the no. repeating thrice are
>> unequal,
>> > > there must exist some bit 'k' such that x[k]!=y[k]. There may be more
>> than
>> > > such bits in x and y. But one such bit certainly yields T[k]=1  after
>> x^y)
>> >
>> > > 2> Now traverse along each bit of  T( in binary) from left or right
>> and
>> > > consider  T[i] =1 which is encountered first. store it . let b=i;
>> > >  (O(M) time and O(M) space to store binary  if M is the bit length of
>> T.)
>> >
>> > > 3> T1= XOR(all elements in given array having bit b as 1)..... (O(N)
>>  time
>> > > and O(M) space) ( time is O(MN) but as M<=32 , complexity remain O(N))
>> >
>> > > 4> T0= XOR( all elements in given array having bit b as 0) (O(N) time
>> and
>> > > O(M) space)
>> >
>> > >  One of (T1,T0) gives the no. that repeats once and the other gives
>> the no
>> > > that repeats thrice.
>> >
>> > > 6> Now traverse the along array A and compute count for T1 and T0. The
>> > > count that equals 3 gives the corresponding no. repeating thrice.
>> -O(N)
>> >
>> > >  Time complexity is O(N+M) . Linear
>> > > space complexity is O(M) to store binary form.
>> >
>> > > But this algo certainly fails if  more than one no. repeats once.
>> >
>> > --
>> > Thanks & Regards,
>> > Priyanka Chatterjee
>> > Final Year Undergraduate Student,
>> > Computer Science & Engineering,
>> > National Institute Of Technology,Durgapur
>> > Indiahttp://priyanka-nit.blogspot.com/- Hide quoted text -
>> >
>> > - Show quoted text -
>>
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>
>
> --
> With Regards,
> Jalaj Jaiswal
> +919026283397
> B.TECH IT
> IIIT ALLAHABAD
>
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