if there are large number of duplicates then it takes long time...
so we have to apply Binary search on rightside.

On Sun, Mar 1, 2009 at 3:35 PM, Prakhar Jain <[email protected]> wrote:

>
> I guess the Soln given is O(n)
> Much better can be achieved through binary serach O(nlogn)
>
> int binsearch(int a[],int start,int end,int key)
> {
> int mid=(start+end)/2;
> if(start==end && a[mid]!=key)
>   return -1;
> if(a[mid]>key)
>    end=mid;
> else if(a[mid]<key)
>    start=mid;
> else
>    return mid;
> }
>
> int search(int a[],int key,int n)
> {
> int p=binsearch(a,0,n,key)
> if(p>=0)
>   while(a[p]==key)
>         p++;
> return p-1;
> }
>
> On Sun, Mar 1, 2009 at 3:14 PM, Miroslav Balaz <[email protected]>
> wrote:
> >
> > cout << ((int)((int*)upperbound(a,a+5,k)-a))/4-1;
> >
> > 2009/3/1 sharad kumar <[email protected]>
> >>
> >> #include<iostream.h>
> >> int main()
> >> {
> >> int a[5]={3,3,3,6,7};
> >> int k;
> >> cin>>k;
> >> int i=0,j=1,c=0;
> >> for(;i<5;i++)
> >> {
> >> if(a[j-1]!=a[i]&& a[i]!=k)
> >> j++;
> >> else
> >> c=i;
> >> }
> >> cout<<c;
> >> return 0;
> >> }
> >>
> >> On Sun, Mar 1, 2009 at 9:50 AM, sharad kumar <[email protected]>
> >> wrote:
> >>>
> >>> hi,
> >>> does the above solution need any time complexity and space complexity
> in
> >>> specifific. idont understand use binary search
> >>> On Sun, Mar 1, 2009 at 12:13 AM, jaanu <[email protected]>
> wrote:
> >>>>
> >>>> Given a sorted arrays of N integers, possibly with duplicates, write a
> >>>> function that
> >>>> returns the highest index of an element X in that array if found or -1
> >>>> otherwise.(use Binary search)
> >>>>
> >>>>
> >>>
> >>
> >>
> >>
> >
> >
> > >
> >
>
>
>
> --
> Prakhar
>
> >
>


-- 




----------------------------------------
CHERUVU JAANU REDDY
M.Tech in CSIS

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