@Amol according to my algo
group 1=9 4 3
group 2= 7 8 1

Think again....

On Sat, Jul 30, 2011 at 6:27 PM, Gary Drocella <gdroc...@gmail.com> wrote:

> To Solve This Problem, I would
> 1. Sort the given list S by their respective strengths.
> 2. Then I would create two other lists A and B for respective
> partitions.
> 3. (a) Remove First and Last from S add them both to A
>    (b) Remove First and Last from S add them both to B
> 4. Repeat Step 3 until there is 1 or 0 people left in which if there
> is 1 person left we would print NO
> 0 people we successfully partitioned the teams into equal strengths.
>
> This is just off the top of my head though, so not sure if it will
> completely work :)
>
> On Jul 30, 8:37 am, Amol Sharma <amolsharm...@gmail.com> wrote:
> > @saurabh- your algo has very high probability of failure
> >
> > take the case  9,7,8,4,3,1
> >
> > acc to ur algo
> > group 1 is  9,8,3  strength =20
> > group 2 is  7,4,2  strength =13
> >
> > but it is possible to divide them into 2 equal grp's
> > take
> > G1 - 9,4,3  total =16
> > G2 - 7,8,1  total =16
> >
> > so we have to think of some better algo
> > --
> >
> > Amol Sharma
> > Third Year Student
> > Computer Science and Engineering
> > MNNIT Allahabad
> >
> >
> >
> >
> >
> >
> >
> > On Sat, Jul 30, 2011 at 5:51 PM, shubham <shubh2...@gmail.com> wrote:
> > > hey sylvester,
> > > just clarify the problem ..
> >
> > > Is it such that in forming the group some people can be left out
> > > or
> > > the sum of the number of people in both partitions is equal to the
> total
> > > number of people
> >
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-- 
Saurabh Singh
B.Tech (Computer Science)
MNNIT ALLAHABAD

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