@ Snehal, Yes you are right, I wrote function for finding max sub array. It can be converted to find min sub array by just switching the comparison operator :)
On Fri, Feb 4, 2011 at 12:14 AM, snehal jain <[email protected]> wrote: > @ Above > u r finding maximum sum subarray > > whereas question is asking for minimum > > On Thu, Feb 3, 2011 at 11:34 PM, Rajiv Podar <[email protected]>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 < >> [email protected]> 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 <[email protected]> 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 <[email protected]> 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 < >>> > >>> > > [email protected]> 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 <[email protected]> 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 < >>> > >>> > > > > [email protected]> 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, 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