@ALL O(n^2 lg(n^2))
http://www.spoj.pl/problems/SUMFOUR/ my code : http://ideone.com/kAPNB plz. suggest some test case's : On Sat, Jun 23, 2012 at 6:59 AM, Amol Sharma <amolsharm...@gmail.com> wrote: > @bhaskar,rammar: > > I don't think your algo willn not work for the following test case -- > > > test case : > arr : 2 4 6 8 > arr^2 : 6 8 10 10 12 14 (sum of each unique pair in arr[i]) > > let's say target sum is 26 > > your solution will return true as they 12+14=26 but in 12 and 14, 8 is > common, infact 26 is not possible in the given array > > -- > > > Amol Sharma > Final Year Student > Computer Science and Engineering > MNNIT Allahabad > > <http://gplus.to/amolsharma99> > <http://twitter.com/amolsharma99><http://in.linkedin.com/pub/amol-sharma/21/79b/507><http://www.simplyamol.blogspot.com/><http://facebook.com/amolsharma99> > > > > > > > On Fri, Jun 22, 2012 at 11:45 PM, Bhaskar Kushwaha < > bhaskar.kushwaha2...@gmail.com> wrote: > >> We first compute the N^2 two sums, and sort the two sums. The for each >> TwoSum t, we check whether there is another two sum t' such that t.value + >> t'.value = target. The time complexity of this approach is O(N^2 logN) >> >> >> On Wed, Jun 20, 2012 at 1:36 AM, rammar <getam...@gmail.com> wrote: >> >>> Lets see ur example... We can have two other arrays corresponding to our >>> n^2 array. >>> For every (target-arr[i]) which we search in our look up array, we can >>> also search the components which were used to get that sum. This can be >>> done in addition constant amount search. >>> I hope we can still go with Hemesh's algo. Please let me know if it >>> breaks somewhere... >>> >>> let's take a test case : >>> arr : 2 4 6 8 >>> arr[0] : 6 8 10 10 12 14 >>> arr[1] : 2 2 2 4 4 6 >>> arr[2] : 4 6 8 6 8 8 >>> >>> >>> P.S. Can we do better? >>> >>> On Wednesday, June 20, 2012 12:22:52 AM UTC+5:30, Amol Sharma wrote: >>>> >>>> @rammar: >>>> can you please explain the case...which i took in the earlier >>>> post..with this method. >>>> >>>> -- >>>> >>>> >>>> Amol Sharma >>>> Final Year Student >>>> Computer Science and Engineering >>>> MNNIT Allahabad >>>> >>>> <http://gplus.to/amolsharma99> >>>> <http://twitter.com/amolsharma99><http://in.linkedin.com/pub/amol-sharma/21/79b/507><http://www.simplyamol.blogspot.com/><http://facebook.com/amolsharma99> >>>> >>>> >>>> >>>> >>>> >>>> >>>> On Tue, Jun 19, 2012 at 11:27 PM, rammar <getam...@gmail.com> wrote: >>>> >>>>> @Hemesh +1 >>>>> >>>>> Please correct me if i am wrong. >>>>> Creation of our look up array a[n*n] -> sum of all the pairs will >>>>> take O(n^2). >>>>> Search using binary sort or quick sort in O(n^2 log (n^2) ) == >>>>> O(n^2 log n) >>>>> We will traverse this array, and for every element we will find >>>>> (target - a[i]) -> This traversal will again take O(n^2). >>>>> For every (target -a[i]) we will search it in our lookup >>>>> array using binary search -> This will take O(log n^2) = O(2log n) = O(log >>>>> n) >>>>> We will store all the matched for the target. >>>>> >>>>> Final complexity = O(n^2) + O(n^2 log n) + O(n^2)*O(log n) == O (n^2 >>>>> log n) >>>>> If the values of max of a[n] is not very high, we can go with a hash >>>>> map. This will result in a quick look up. And we can get the answer in >>>>> O(n^2). >>>>> >>>>> >>>>> P.S. Can we do better? >>>>> >>>>> >>>>> On Monday, June 18, 2012 6:10:33 PM UTC+5:30, Jalaj wrote: >>>>>> >>>>>> @KK and hemesh >>>>>> target is not a constant value , it can be any element in array , so >>>>>> you need to do binary search for all (array[i] - (a+b)) to find which >>>>>> increases the complexity to n^3logn. >>>>>> So, i think the n^3 approach which i gave before do it ?? >>>>>> >>>>>> ------ Correct me if m wrong >>>>>> >>>>>> On Mon, Jun 18, 2012 at 2:58 PM, Amol Sharma >>>>>> <amolsharm...@gmail.com>wrote: >>>>>> >>>>>>> @hemesh,kk: >>>>>>> >>>>>>> let's take a test case : >>>>>>> arr : 2 4 6 8 >>>>>>> arr^2 : 6 8 10 10 12 14 (sum of each unique pair in >>>>>>> arr[i]) >>>>>>> >>>>>>> let's say target sum is 26 >>>>>>> >>>>>>> your solution will return true as they 12+14=26 but in 12 and 14, 8 >>>>>>> is common, infact 26 is not possible in the given array >>>>>>> >>>>>>> can u please elaborate how will you take care of such situation ? >>>>>>> >>>>>>> @jalaj: >>>>>>> yes it's O( (n^3)*logn) >>>>>>> >>>>>>> @bhavesh: >>>>>>> fyi.. >>>>>>> log(n^3)=3*log(n)=O(log(n)) >>>>>>> so it's same.. :P >>>>>>> >>>>>>> >>>>>>> >>>>>>> >>>>>>> >>>>>>> -- >>>>>>> >>>>>>> >>>>>>> Amol Sharma >>>>>>> Final Year Student >>>>>>> Computer Science and Engineering >>>>>>> MNNIT Allahabad >>>>>>> >>>>>>> <http://gplus.to/amolsharma99> >>>>>>> <http://twitter.com/amolsharma99><http://in.linkedin.com/pub/amol-sharma/21/79b/507><http://www.simplyamol.blogspot.com/><http://facebook.com/amolsharma99> >>>>>>> >>>>>>> >>>>>>> >>>>>>> >>>>>>> >>>>>>> >>>>>>> On Mon, Jun 18, 2012 at 12:29 AM, KK <kunalkapadi...@gmail.com>wrote: >>>>>>> >>>>>>>> @Hemesh : +1 >>>>>>>> @Jalaj : read Hemesh's solution again it is for 4sum. >>>>>>>> In short, make a new array having sum of each unique pair of given >>>>>>>> array. -> O(n^2) >>>>>>>> sort it -> O(n^2) >>>>>>>> for each number bi in new array, binary search (target - bi) in the >>>>>>>> same array -> O(n^2) >>>>>>>> >>>>>>>> >>>>>>>> On Sunday, 17 June 2012 12:41:40 UTC+5:30, Jalaj wrote: >>>>>>>>> >>>>>>>>> The solution which hemesh gave was solution to 3SUM hard problem >>>>>>>>> the best solution for which can be achieved in n^2 . >>>>>>>>> And the original question is a kind of 4SUM hard problem for which >>>>>>>>> best possible solution i think is again n^3 and Amol what you told is >>>>>>>>> not >>>>>>>>> n^3 , finding all triplets will itself take n^3 and doing a binary >>>>>>>>> search >>>>>>>>> again that sums upto n^3*logn. >>>>>>>>> >>>>>>>>> @shashank it is not a variation of 3SUM problem as in 3SUM problem >>>>>>>>> a+b+c = some constant , but in your case it is "b+c+d = s-a", where a >>>>>>>>> can >>>>>>>>> change again and again so if you do even apply 3SUM logic to it you >>>>>>>>> will >>>>>>>>> have to do it for every a which will make it n^2*n = n^3 >>>>>>>>> >>>>>>>>> >>>>>>>>> >>>>>>>>> On Sat, Jun 16, 2012 at 2:45 AM, sanjay pandey < >>>>>>>>> sanjaypandey...@gmail.com> wrote: >>>>>>>>> >>>>>>>>>> @hemesh cud u plz elaborate wat is b[k]=a[i]+a[j]...n also ur >>>>>>>>>> solution... >>>>>>>>>> >>>>>>>>>> -- >>>>>>>>>> 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+unsubscribe@**googlegr****oups.com<algogeeks%2bunsubscr...@googlegroups.com> >>>>>>>>>> . >>>>>>>>>> For more options, visit this group at http://groups.google.com/** >>>>>>>>>> group****/algogeeks?hl=en<http://groups.google.com/group/algogeeks?hl=en> >>>>>>>>>> . >>>>>>>>>> >>>>>>>>> >>>>>>>>> >>>>>>>>> >>>>>>>>> >>>>>>>>> -- >>>>>>>> You received this message because you are subscribed to the Google >>>>>>>> Groups "Algorithm Geeks" group. >>>>>>>> To view this discussion on the web visit >>>>>>>> https://groups.google.com/d/**ms**g/algogeeks/-/9jCCN5iHDB8J<https://groups.google.com/d/msg/algogeeks/-/9jCCN5iHDB8J> >>>>>>>> . >>>>>>>> >>>>>>>> To post to this group, send email to algogeeks@googlegroups.com. >>>>>>>> To unsubscribe from this group, send email to >>>>>>>> algogeeks+unsubscribe@**googlegr**oups.com<algogeeks%2bunsubscr...@googlegroups.com> >>>>>>>> . >>>>>>>> For more options, visit this group at http://groups.google.com/** >>>>>>>> group**/algogeeks?hl=en<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+unsubscribe@ >>>>>>> **googlegr**oups.com <algogeeks%2bunsubscr...@googlegroups.com>. >>>>>>> For more options, visit this group at http://groups.google.com/** >>>>>>> group**/algogeeks?hl=en<http://groups.google.com/group/algogeeks?hl=en> >>>>>>> . >>>>>>> >>>>>> >>>>>> >>>>>> >>>>>> -- >>>>>> Regards, >>>>>> >>>>>> Jalaj Jaiswal >>>>>> Software Engineer, >>>>>> Zynga Inc >>>>>> +91-9019947895 >>>>>> * >>>>>> * >>>>>> FACEBOOK <http://www.facebook.com/jalaj.jaiswal89> >>>>>> LINKEDIN<http://www.linkedin.com/profile/view?id=34803280&trk=tab_pro> >>>>>> >>>>>> -- >>>>> You received this message because you are subscribed to the Google >>>>> Groups "Algorithm Geeks" group. >>>>> To view this discussion on the web visit https://groups.google.com/d/* >>>>> *msg/algogeeks/-/SGN_A_YrZlkJ<https://groups.google.com/d/msg/algogeeks/-/SGN_A_YrZlkJ> >>>>> . >>>>> >>>>> To post to this group, send email to algogeeks@googlegroups.com. >>>>> To unsubscribe from this group, send email to algogeeks+unsubscribe@** >>>>> googlegroups.com <algogeeks%2bunsubscr...@googlegroups.com>. >>>>> For more options, visit this group at http://groups.google.com/** >>>>> group/algogeeks?hl=en <http://groups.google.com/group/algogeeks?hl=en> >>>>> . >>>>> >>>> >>>> -- >>> You received this message because you are subscribed to the Google >>> Groups "Algorithm Geeks" group. >>> To view this discussion on the web visit >>> https://groups.google.com/d/msg/algogeeks/-/eDvKXozaZV8J. >>> >>> 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. >>> >> >> >> >> -- >> regards, >> Bhaskar Kushwaha >> Student >> CSE >> Third year >> M.N.N.I.T. Allahabad >> >> >> -- >> 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. >> > > -- > 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. > -- 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. 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