@Dave..:-O well catch...

On Sat, Sep 3, 2011 at 9:00 PM, Dave <[email protected]> wrote:

> @Piyush: Try your code with
>
> n = 10
> a[10] = {11,22,33,44,55,66,77,88,99,110}
> b[10] = {10,20,30,40,50,60,70,80,90,100}
>
> Your code gets
>
> (110, 100) = 210
> (110, 90) = 200
> (99, 100) = 199
> (110, 80) = 190
> (88, 100) = 188
> (110, 70) = 180
> (77, 100) = 177
> (110, 60) = 170
> (66, 100) = 166
> (110, 50) = 160
>
> but it should get
>
> (110, 100) = 210
> (110, 90) = 200
> (99, 100) = 199
> (110, 80) = 190
> (99, 90) = 189
> (88, 100) = 188
> (110, 70) = 180
> (99, 80) = 179
> (88, 90) = 178
> (77, 100) = 177
>
> It fails because, after choosing the first four pairs, it does not
> consider all three candidates, (110, 70) = 180, (99, 90) = 189, and
> (88, 100) = 188. It only looks at the first and last of these. Later
> on, it fails to consider (99, 80) = 179 and (88, 90) = 178.
>
> After you have chosen the maximum pair, every unused pair that is the
> last unused pair in both its row and column of the (implicit) n by n
> matrix of pairwise sums is a candidate. When you have chosen n-1
> pairs, there can be O(sqrt(n)) candidates for the last choice. You are
> considering only two of them.
>
> Dave
>
> On Sep 2, 3:09 pm, Piyush Grover <[email protected]> wrote:
> > if I have understood the question correctly then:
> >
> > a[n-1] + b[i] > a[j] + b[i] for all 0 <= j <  n-1
> > and a[j] + b[n-1] > a[j] + b[i] for all 0 <= i < n-1
> > therefore,
> >
> > i = j =n-1;
> > count = 1;
> > S[0] <-- (a[n-1], b[n-1])
> > p = a[n-1] + b[n-2];
> > q = a[n-2] + b[n-1]
> >
> > while(count < n){
> >
> >     if(p > q){
> >          j--;
> >          S[count++] <-- (a[n-1], b[j]);
> >     }else{
> >         i--;
> >         S[count++]  <-- (a[i], b[n-1]);
> >     }
> >
> >     p = a[n-1] + b[j-1];
> >     q = a[i-1] + b[n-1];
> >
> > }
> >
> > Time complexity: O(n)  :  http://ideone.com/FXfVj
> >
> > On Fri, Sep 2, 2011 at 10:05 PM, WgpShashank <[email protected]
> >wrote:
> >
> >
> >
> > > @Dave Correct , Missed to Provide the Correct Time Complexity in Worst
> Case
> > > it Will be O(N^2) , as we need to find out n  such maximum pair , will
> think
> > > about O(N0) Algo, if able to do it, will post the Algo here
> >
> > > Thanks
> > > Shashank Mani
> > > Computer Science
> > > Birla Institute of Technology,Mesra
> >
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