If we do this we cannot achieve O(n) running time.
Here we can access only one bit at a time.
Abhijit

On 5/14/07, Kumaraguru Paramasivam <[EMAIL PROTECTED]> wrote:
>
> Find the sum of all the numbers in the array.[s1]
>
> Since the array contains integers from 0 or 1 to 'n'.
> Find the sum of this first n natural numbers n(n+1)/2. [s2]
>
> Missing number = [s2] - [s1]
>
>
> On 5/12/07, You I <[EMAIL PROTECTED]> wrote:
>
> > Hi Googmeister,
> >
> > You wrote "but the idea easily extends to arbitrary n"
> > Could you explain how ?
> >
> > Thanks,
> > AlgoStudent
> > On Jun 21 2006, 9:43 pm, "Googmeister" <[EMAIL PROTECTED]> wrote:
> > > anil kumar wrote:
> > > > An array A[1..n] contains all the integers from 0 to n except one.
> > It
> > > > would be easy to determine themissingintegerin O(n) time by using an
> > > > auxilary array B[0..n] to record which numbers appear in A. In this
> > > > problem however we cannot access an entireintegerin A with a single
> > > > operation. The elements of A are represented in binary, the only
> > > > operation we can use to access them is " Fetch the jth bit of A[i] "
> > ,
> > > > which takes constant time.Findthemissingintegerin O(n) time using
> > > > only that operation.
> > >
> > > Are you permitted to swap array entries in constant time?
> > > If so, the following is a solution. I'll assume n is a power of 2
> > > for simplicity (but the idea easily extends to arbitrary n).
> > >
> > > Scan through the leading bits of the n integers.
> > Themissingintegerstarts with 0 if 0 appears an odd number of times,
> > > and 1 otherwise. Move all the integers starting with the same
> > > leading bit as themissingintegerto one side of the array
> > > (e.g., ala partitioning in quicksort). Now recur on those
> > > remaining integers and the next most significant bit. There
> > > are lg n phases since the number of bits perintegeris lg n,
> > > but the overall running time is still linear: n + n/2 + n/4 + 
> > > ...<[email protected]>
> > > >
> >

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