@ Above u r finding maximum sum subarray whereas question is asking for minimum
On Thu, Feb 3, 2011 at 11:34 PM, Rajiv Podar <rajeevpo...@gmail.com> wrote: > Hi Richi, > > Thanks for finding the problem. I forgot to add one condition: > Please have a look on the following code. I hope this will solve the > problem. > > int array[] = {-1,-2,3,4}; > int length = 4; > > int max = 0; > int current = 0; > > for (int i = 0; i < length; i++) > { > current += array[i]; > if (current < 0 ) > { > current = 0; > } > max = max > current ? max : current; > } > std::cout<<"Max is : "<<max; > > > > Thanks & Regards, > Ricky > > > > On Thu, Feb 3, 2011 at 6:58 PM, sankalp srivastava < > richi.sankalp1...@gmail.com> wrote: > >> @above guy with cheers and all and snehal >> >> the best way to prove wrong is by a test case , so , >> >> -1 -2 3 4 >> >> Ricky's solution will give the answer as 4 , while the answer is 7 >> >> @snehal . >> [quote]if indices starting at 1 >> bothers you then take >> >> P[i]= A[0] + A[1] + .... + A[i] >> its one and the same thing.. [\Quote] >> I'm really not that stupid to bother about an off-by-one error :-) >> Your algo rephrased :- >> P[i] = A[1] + A[2] + … + A[i] >> so , >> P[1]=-1 >> P[2]=-3 >> P[3]=0 >> P[4]=4 >> >> Q[i].Value = P[i]. >> Q[i].Index = i >> >> So , >> >> Q[1]=-1 , 1 >> Q[2]=-3 , 2 >> Q[3]=0 , 3 >> Q[4]=4 , 4 >> >> Now , as u said , let's sort it >> >> new Q={{4 , 4} ,{0 , 3} ,{-1 , 1} ,{-3 , 3}} >> You din mention anything after this , so I dunnoe what you plan up >> from here . How are we going to get the answer {3 , 4 } from here ? >> >> Now , >> >> >> On Feb 2, 10:06 pm, Ricky <rajeevpo...@gmail.com> wrote: >> > Hi you can try the following code snippet: >> > int array[] = {11, -12, 15, -3, 8, -9, 1, 8, 10, -2}; >> > int length = 10; >> > >> > int max = 0; >> > int current = 0; >> > >> > for (int i = 0; i < length; i++) >> > { >> > current += array[i]; >> > max = max > current ? max : current; >> > } >> > std::cout<<"Max is : "<<max; >> > >> > Cheers!!!!!!!!!! >> > >> > On Feb 2, 9:04 pm, snehal jain <learner....@gmail.com> wrote: >> > >> > >> > >> > > @ above >> > > didnt get you? why is the solution wrong? and if indices starting at 1 >> > > bothers you then take >> > >> > > P[i]= A[0] + A[1] + .... + A[i] >> > > its one and the same thing.. >> > >> > > On Wed, Feb 2, 2011 at 6:02 PM, sankalp srivastava < >> > >> > > richi.sankalp1...@gmail.com> wrote: >> > >> > > > This solution is wrong , never has it been said that the indices >> will >> > > > occur from 1.....i (if that is the case , you don't need to sort , >> > > > just return the maximum /minimum sum obtained as a result) >> > >> > > > The indices were b/w i..j such that 1<=i<=j<=n >> > >> > > > O(n) solution does not exist .The state space tree will have n! leaf >> > > > nodes(because there is some ordering on the input data , otherwise >> it >> > > > would have 2^n leaf nodes) .Traversing the tree will take O(log n!) >> > > > steps , or O(n log n) >> > > > In fact a slight modification to this , the subset sum problem id >> NP- >> > > > complete . >> > > > But with the Q[i] array , you can get the answer with simple >> recursion >> > > > ( or bfs or state space search or dp ) . >> > >> > > > On Feb 2, 1:33 pm, snehal jain <learner....@gmail.com> wrote: >> > > > > @ 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 >> > > > > > 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