Hi,
I have understood the solution, but here we are talking about knowing the
path(elements in the subsets in this case).
I saw we could use the back pointer technique, which I understood, but I'm
not able to see how would I code this technique.
Please try to explain me this thing.
Thanks a lot in advance.
On Wed, Jan 9, 2013 at 10:46 AM, Don <dondod...@gmail.com> wrote:
> My solution is equivalent to using DP for the 0-1 knapsack, but the DP
> approach does not identify the partition, it only determines if it
> exists. In the same way, my solution does not determine which numbers
> to make negative. It only determines what the smallest possible sum
> is. The DP approach to 0-1 knapsack is not really a solution, if the
> problem is stated in the usual way: what set of objects should I put
> in the knapsack to maximize the total. It only solves the problem:
> what is the maximum total I can achieve.
>
> On Jan 9, 7:12 am, Carl Barton <odysseus.ulys...@gmail.com> wrote:
> > How is it a huge improvement? Your O(SN) is the same time as the dynamic
> > programming solution for 0-1 knapsack isn't it?
> >
> > On 8 January 2013 14:44, Don <dondod...@gmail.com> wrote:
> >
> >
> >
> >
> >
> >
> >
> > > Yes, that is true. However, trying to find the optimal partition is
> > > equivalent to the 0-1 knapsack problem, which is exponential time. So
> > > S*N is a huge improvement over that. Does someone have a better
> > > solution?
> > > Don
> >
> > > On Jan 7, 10:49 am, Nikhil Karnwal <sunnyk12...@gmail.com> wrote:
> > > > @ Don
> > > > but ur's solution complexity is O(S*N) which is large in case of
> large N
> > > > and large numbers.
> > > > Like in case of s=1000000 and N=10^5.
> > > > Correct me if I am wrong.
> >
> > > > Nikhil Karnwal
> >
> > > > On Mon, Jan 7, 2013 at 9:04 PM, Don <dondod...@gmail.com> wrote:
> > > > > Note that you don't need to store the entire P matrix. You really
> just
> > > > > need the last column.
> > > > > Don
> >
> > > > > On Jan 7, 10:29 am, Don <dondod...@gmail.com> wrote:
> > > > > > You want to partition the array A into to subsets S1 and S2 such
> that
> > > > > > you minimize |Sum(S1)-Sum(S2)|.
> >
> > > > > > The optimal sum for the subsets is S=SUM(A)/2
> >
> > > > > > Use DP to build a matrix P:
> > > > > > P[i][j] = 1 if some subset of {A[0]..A[i]} has a sum of j, 0
> > > otherwise
> >
> > > > > > Now find a value of i such that P[n][i] = 1 which minimizes S-i.
> >
> > > > > > The minimum sum is 2S-2i.
> >
> > > > > > Don
> >
> > > > > > On Jan 5, 12:58 pm, mukesh tiwari <mukeshtiwari.ii...@gmail.com>
> > > > > > wrote:
> >
> > > > > > > Hello All!
> > > > > > > I have a given array of numbers and I have to change the sign
> of
> > > > > numbers to
> > > > > > > find out the minimum sum. The minimum sum will be 0 or greater
> > > than 0.
> > > > > Here
> > > > > > > is couple of test cases
> > > > > > > 1. [ 1 , 2 , 3 , 2 , 4 ]. Changing the sign [ -1 , -2 , -3 ,
> 2 ,
> > > 4 ]
> > > > > so
> > > > > > > minimum sum will be 0.
> > > > > > > 2. [ 3 , 5 , 7 , 11 , 13 ]. Changing the sign [ -3 , -5 , 7 ,
> -11
> > > ,
> > > > > 13 ]
> > > > > > > so minimum sum is 1.
> >
> > > > > > > So technically this problem boils down to divide the set into
> two
> > > > > subset
> > > > > > > and find out the minimum difference. I though of DP but the
> number
> > > of
> > > > > > > element in array could 10^5 so could some one please tell me
> how to
> > > > > solve
> > > > > > > this problem ? I didn't assume that number will be positive
> > > because it
> > > > > was
> > > > > > > not given in the problem.
> >
> > > > > > > Regards
> > > > > > > Mukesh Tiwari
> >
> > > > > --
> >
> > > --
>
> --
>
>
>
--
PRANKUR GUPTA
Masters Student (CSE)
State University of New York
Stony Brook University
prgu...@cs.stonybrook.edu
--
