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 <[email protected]> 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 <[email protected]> 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 <[email protected]> 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 <[email protected]> > > > 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 > > > -- --
