@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 < [email protected]> 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 <[email protected]> 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 <[email protected]> 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 >>>>> <[email protected]>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 <[email protected]>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 < >>>>>>>> [email protected]> 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 [email protected]. >>>>>>>>> To unsubscribe from this group, send email to >>>>>>>>> algogeeks+unsubscribe@**googlegr****oups.com<algogeeks%[email protected]> >>>>>>>>> . >>>>>>>>> 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 [email protected]. >>>>>>> To unsubscribe from this group, send email to algogeeks+unsubscribe@ >>>>>>> **googlegr**oups.com <algogeeks%[email protected]>. >>>>>>> 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 [email protected]. >>>>>> To unsubscribe from this group, send email to algogeeks+unsubscribe@* >>>>>> *googlegr**oups.com <algogeeks%[email protected]>. >>>>>> 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 [email protected]. >>>> To unsubscribe from this group, send email to algogeeks+unsubscribe@** >>>> googlegroups.com <algogeeks%[email protected]>. >>>> 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 [email protected]. >> To unsubscribe from this group, send email to >> [email protected]. >> 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 [email protected]. > To unsubscribe from this group, send email to > [email protected]. > 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 [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
