@ankit sinha:i dont think the algo u wrote is o(logn).....its o(N) i guess
On Thu, Mar 3, 2011 at 7:05 PM, Ankit Sinha <akki12...@gmail.com> wrote:
> @sukhmeet, as per the question, there is a unique value which satisfy
> a[i] = i in the array and written code captures that only. Else we
> need to modify if we are interested in all such values which statisfy
> the said condition. And also in that case looping around bsearch will
> not be a good idea.. it will be better to go for simple loop in o(n)
> time as every time mid calculation is also costly. I agreed to Arpit
> view point and accordingly did coding as preetika needed as per arpit
> comments.
>
> Cheers!!
> Ankit
>
> On Thu, Mar 3, 2011 at 6:53 PM, sukhmeet singh <sukhmeet2...@gmail.com>
> wrote:
> > what should be the answer for this:
> > if A={0,1,2,4,5}
> > 0 or 1 or 2
> > On Thu, Mar 3, 2011 at 6:26 PM, Ankit Sinha <akki12...@gmail.com> wrote:
> >>
> >> Hi,
> >>
> >> Here is the code to do this using Bsearch in o(logn) time.
> >>
> >> int BsearchElemEqualIndex (int *a, int start, int end)
> >> {
> >> int mid = (((end - start) >> 1) + start);
> >> if (a[mid] == mid)
> >> return a[mid];
> >> else if (a[mid] != mid)
> >> {
> >> if (mid == start || mid == end)
> >> {
> >> return -1;
> >> }
> >> else
> >> {
> >> BsearchElemEqualIndex (a, start, mid);
> >> BsearchElemEqualIndex (a, mid + 1, end);
> >> }
> >> }
> >> }
> >>
> >> int _tmain(int argc, _TCHAR* argv[])
> >> {
> >> int a[SIZE] = {5,9,3,8,1,2,6,7};
> >> int x = BsearchElemEqualIndex (a, 0, SIZE);
> >> printf ("%d", x);
> >> system ("PAUSE");
> >> return 0;
> >> }
> >>
> >> Cheers,
> >> Ankit!!!
> >>
> >> On Thu, Mar 3, 2011 at 11:04 AM, Param10k <paramesh...@gmail.com>
> wrote:
> >> > There is a sorted array and you have to write a fn to find a number in
> >> > the array which satisfies
> >> >
> >> > A[i] = i ; where A is the array and i is the index...
> >> >
> >> > - Param
> >> > http://teknotron-param.blogspot.com/
> >> >
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