@Arun
Elegant!
I see it like a reduction of Integer partition problem to Knapsack
problem.
Where:-
The target is SUM_OF_GIVEN_INTEGERS/2 and we want to get as close to
the target as possible.

_dufus


On Aug 15, 8:31 pm, Arun N <arunn3...@gmail.com> wrote:
> This can be solved in single dimension itself
>
> like this
>
> SUM = sum of elements
> m = SUM/2
>
> DP[0] =1;   // DP[ i ]  = 1 if it is possible to reach sum ' i ' using the
> elements
>
> for ( i=0 ; i<n ; i++ )
>     for ( j = m ; j>=0 && j >=a[i]; j-- )
>         DP[j] = DP[j-a[i]];
>
> for( j = m ; j>=0 ; j--)
>         if(DP[j]) break;
>
> so the answer is
>
> j and SUM - j
>
> Arun,
>
>
>
> On Sat, Aug 15, 2009 at 1:02 PM, Dufus <rahul.dev.si...@gmail.com> wrote:
>
> > Plz refer to
> > Balanced Partition Problem or
> > Integer Partition Problem at
>
> >http://people.csail.mit.edu/bdean/6.046/dp/
>
> > On Aug 14, 10:26 pm, fundoonick <fundoon...@yahoo.co.in> wrote:
> > > @DufusCan u pls give the algorithm about how to do this?
>
> > > On Fri, Aug 14, 2009 at 8:56 PM, Dufus <rahul.dev.si...@gmail.com>
> > wrote:
>
> > > > If the range is given then it get reduced to a standard problem which
> > > > can be solved by DP in O(k.n^2) time..where n integers within range
> > > > 0...K have to be partitioned in two sets S1 and S2 such that the
> > > > difference of sum of their elements is min.
>
> > > > _dufus
>
> > > > On Aug 14, 6:27 pm, fundoonick <fundoon...@yahoo.co.in> wrote:
> > > > > Problem:
> > > > > I have a set of positive integers. I have to divide it into 2 sets
> > such
> > > > that
> > > > > the difference of the sums of both sets is minimum.
> > > > > For ex, the given set of +ve integers is: 1,2,3,4
> > > > > I divide it into 2 sets {1,4} and {2,3} such that the difference of
> > their
> > > > > sum (1+4=)5 - (2+3=)5 = 0
> > > > > This is the least possible difference.
>
> > > > > Pls help.
>
> --
> Potential is not what U have, its what U think U have!!!
> It is better to worn out than rust.

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