@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@** >>>>> 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/-/9jCCN5iHDB8J. >>> >>> 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. >> > > > > -- > 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. 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.