sumedh, your solution is good but dsha did not ask about it,
if u read question carefully then notice term using constant memory or space
but in your solution space complexity will be very high.
Is there a way to find this element faster than
O(n*log n) and with constant extra memory?
so plz give ans according to it.
On 8/17/07, sumedh sakdeo <[EMAIL PROTECTED]> wrote:
>
> arr[]={1,2,4,5,67,89,4}
> o(n) storing time into hashtable
> index cnt
> 0 0
> 1 1
> 2 1
> 3 0
> 4 1(if(h[4] ==0) then enter else dont
> 5 1
> ... 0
> ... 0
> 67 1
> 89 1
>
> check before entering into the hash table if the value at that index is 0?
> O(1)
> if it is then the number has ocured once..increment value in h[i]
> if not, then number is occuring for second time..
> space required is more...to store table..
> but space is cheaper than time...
>
> Sumedh
>
> On 8/17/07, Vaibhav Jain <[EMAIL PROTECTED]> wrote:
> >
> > if u know the range of values stored in array then
> > let me assume values 1 to k then u can calculate sum of numbers stored
> > in array in O(n) complexity.
> > after that apply formula
> >
> > duplicate value= [k*(k+1)]/2 - sum of numbers stored in array
> >
> > it will take O(1) constant time so total complexity becomes only O(n).
> >
> > it can be one solution to your problem but if the range is unknown for
> > values then
> > is there any solution to come in O(n)???
> >
> >
> >
> > On 8/16/07, dsha <[EMAIL PROTECTED]> wrote:
> > >
> > >
> > > Hi there,
> > >
> > > I'm interested in the following problem: there is an array of integers
> > >
> > > that contains each element only once except for one element that
> > > occurs exactly twice. Is there a way to find this element faster than
> > > O(n*log n) and with constant extra memory? If no, how can I prove it?
> > >
> > > Thanks in advance for ideas.
> > >
> > >
> > >
> > >
> > >
> >
> >
> > --
> > Vaibhav Jain
> > > >
> >
--
Vaibhav Jain
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