@ above its nt any homework question.. i found it a good question... aftr spending a lot of time i came up with following solution
Given Input Array A form the prefix sum array P of A. i.e P[i] = A[1] + A[2] + … + A[i] Now create another array Q of pairs (Value, Index) such that Q[i].Value = P[i]. Q[i].Index = i Now sort that array Q, considering only Q[i].Value for comparison. We get a new sorted array Q’ such that Q’[i].Value Q’[i].Index Time complexity o( nlogn) and my O(n) which i posted earlier is giving incorrect result in some case..so ignore that.. so does there exist O(n) solution for it also.. i had tried a lot but could not figure out. but i think it should exist as there is for the other variation.. On Tue, Feb 1, 2011 at 8:24 PM, sankalp srivastava < richi.sankalp1...@gmail.com> wrote: > > You should not post homework problems . > 1)For divide and conquer :- > Read about interval trees , binary indexed trees , segments > trees . > Solve this using interval tree (By the time you solve a few > basic problems of interval tree , you would be able to figure out a > solution) > > > the function to calculate the parent will be > 1) first check if the two are +ve > 2) if yes , join both of them and also iterate on the sides left by > both , to see if you can include them also (You only need to see the > positive elements , no negative elements ) > > T(n)=2T(n/2)+O(n) > > I gan explain in detail , please correct me if im wrong > > Logic :- Basically in the subproblem , we would have founded the > maximum subarray in that well , subarray (short of words ) .So , if we > want to ,we can only increase the solution in the next subarray (the > second subproblem ) > So , there will be three cases > > Either the subarray , the most minimum sum in one of the subproblems > will be the answer > The answer will be from between the gap of the indices between the > solutions of the two subproblems > The answer will be any combination of the two > > All these three can be checked in O(n) itself . > > 2)Using DP(I don't know how many dp (pure dp i mean) algorithms are in > O(nlog n) .Never heard of any with the pure dp approach and an n log n > solution ) > > DP(classical for maximum positive sum array ) can be done by going > through two loops > > dp[i]= minimum positive sum for an array with index (last index =i ) > p[i]= start index corresponding to this dp[i] > > dp[i]= minimum sum condition ( for each i<j ) > update p[i] accordingly .Then return the minimum amongst dp[i] and > corresponding p[i] . > > This is a complete search , so I don't think it will get wrong . > > And i don't think it could be solved in O(n log n) (at least with > dp) .Because the search space tree would be of height O(log n) (with > no overlapping problems ) and dp lives upon overlapping subproblems . > Or may be , if someone could provide with a O( n log n) solution > > Regards , > Sankalp Srivastava > > "I live the way I type , fast and full of errors " > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algogeeks@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups.com> > . > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
