hi sourab please explain what bit vector1 and bit vector 2 really are can 
you give an example please?

On Saturday, February 16, 2013 11:20:59 PM UTC+5:30, sourabh wrote:
>
> you can solve this problem using bitvector/bitset.
>
> first scan :
> scan the array set the bit on odd occurrence and unset on even or 
> 0 occurrence.
>
> second scan :
> shift all the odd occurring elements in beginning of array and even 
> towards end.
>
> third scan : till end of odd occurring elements.
> take another bit vector 
> on first occurence :if bit is set in bv1 then unset it and set it in bv2.
> on second occurence : if bv1 is not set and bv2 is set then these are the 
> array elements occuring 3rd time. 
>
>
>
>
>
> On Wed, Feb 13, 2013 at 9:27 PM, prakhar singh 
> <[email protected]<javascript:>
> > wrote:
>
>> Yes, thats a valid point Don.
>> Thats what i meant when i wrote  "//is that correct?" in the comments on 
>> the array line in code.
>>
>>
>>  int a[] = {2,2,3,3,3,1,1,4,4};   // is this correct?
>>
>>
>> On Wed, Feb 13, 2013 at 9:09 PM, Don <[email protected] <javascript:>>wrote:
>>
>>> The xor approach only works if there are no values which occur only
>>> once. But the problem statement indicates that some numbers occur
>>> once, some occur twice, and one occurs three times. So you will end up
>>> with prod equal to the xor of all of the values which occur once or
>>> three times. Put that in your input array and you'll find that you
>>> don't get the desired output.
>>>
>>> I don't know of a solution better than sorting and scanning the array.
>>>
>>> Don
>>>
>>> On Feb 12, 3:14 pm, prakhar singh <[email protected]> wrote:
>>> > #include<stdio.h>
>>> >
>>> > int main()
>>> > {
>>> >    int a[] = {2,2,3,3,3,1,1,4,4};   // is this correct?
>>> >    int prod=a[0];int i;
>>> >    for(i=1;i<(int)sizeof(a)/sizeof(a[0]);i++)
>>> >    {
>>> >       prod ^= a[i];
>>> >    }
>>> >    printf("%d\n",prod);   //outputs 3, algorithm works as Sachin 
>>> described
>>> > it;
>>> >
>>> > }
>>> >
>>> > On Tue, Feb 12, 2013 at 11:44 PM, Sachin Chitale
>>> > <[email protected]>wrote:
>>> >
>>> >
>>> >
>>> >
>>> >
>>> >
>>> >
>>> > > use ex-or operation for all array elements..
>>> > > a^a=0
>>> > > a^a^a=a
>>> >
>>> > > On Sun, Feb 10, 2013 at 8:22 AM, Mohanabalan D B <
>>> [email protected]>wrote:
>>> >
>>> > >> can use counting sort
>>> >
>>> > >> On Sun, Jul 15, 2012 at 6:37 PM, santosh thota <
>>> [email protected]>wrote:
>>> >
>>> > >>> If we can retrieve ith prime efficiently, we can do the 
>>> following...
>>> > >>> 1.maintain a prod=1, start from 1st element, say a[0]=n find n th 
>>> prime
>>> > >>> 2.check if (prod% (ith_prime * ith_prime )==0) then return i;
>>> > >>>            else prod=prod*ith_prime;
>>> > >>> 3.repeat it till end
>>> >
>>> > >>> On Thursday, 12 July 2012 10:55:02 UTC+5:30, algo bard wrote:
>>> >
>>> > >>>> Given an array of integers where some numbers repeat once, some 
>>> numbers
>>> > >>>> repeat twice and only one number repeats thrice, how do you find 
>>> the number
>>> > >>>> that gets repeated 3 times?
>>> >
>>> > >>>> Does this problem have an O(n) time and O(1) space solution?
>>> > >>>> No hashmaps please!
>>> >
>>> > >>>  --
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>>> > > Application Engineer SCJP, SCWCD
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>>  
>>  
>>
>
>
>
> -- 
> Regards,
> Sourabh Kumar Jain
> +91-8971547841 
>

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