Sorry it is  some not sum...

Yes ,actually the array is infinite,has some n elements in sorted order ,may
not be continous and if the array has invalid numbers it has -1 in
it.Nowthe task is to find a given  valid
number.Pls help me out now?

On 5/7/07, Nisha Subramanian <[EMAIL PROTECTED]> wrote:
>
> Yes ,actually the array is infinite,has sum n elements in sorted order
> ,may not be continous and if the array has invalid numbers it has -1 in
> it.Now the task is to find a given  valid number.Pls help me out now?
>
> On 5/7/07, Arun <[EMAIL PROTECTED]> wrote:
> >
> > BiGYan,
> > I think ur solution will not work as u r assuming the the "valid"
> > numbers
> > are stored contigously and n>inc
> >
> > > while ( arr[i].val<=s )
> > >    i+=inc;
> >
> > I think the question can be rephrased as a very large array of size N,
> > has n valid numbers (n<<N) which are sorted(need not be contigous).
> > Task is to find if s is present in O(n)
> >
> > I think u wanted to say something like increment step by 'inc', look
> > "around" if u can find a valid value and then step forward or back.
> > This is only sol'n I can think of too. Its only a heuristic, complexity
> > is still O(N),
> > not O(n).
> >
> >
> > On 5/6/07, BiGYaN < [EMAIL PROTECTED]> wrote:
> > >
> > >
> > > Infinite Array ?
> > > Infinite Array in sorted order ?
> > >
> > > Let's say we have an infinite array where each of the elements have a
> > > number (according to which it is sorted) and a valid/invalid bit for
> > > storing the validity.
> > >
> > > s - number we are searching for
> > > arr[] - the infinite array
> > >
> > > inc = 100;    // this is the value you gotta tweak
> > > i=0;
> > > while ( arr[i].val<=s )
> > >     i+=inc;
> > >
> > > if ( i==0 )
> > >     // value not found
> > > else
> > >     binary_search(arr[],i-inc,i,s);    // arr[], low_lim, hi_lim,
> > > search_number
> > >
> > > Guess this is the best that you could get.
> > > But I'm still confused about that "Infinite Array".
> > >
> > >
> > >
> > >
> > >
> >
> > > >
> >
>

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