@fenghuang try this array:
a[]={3,3,3,3,3,3,4,4,4,4,4,4,5}
so asq[]={9,9,9,9,9,9,16,16,16,16,16,16,25}
here as u can see the total number of requisite triple pairs are 6*6=36,
in general for above array total number of pairs is (n/2)*(n/2) i.e. n^2/4
where n is the size of the array.
by using O(n) algo and since you are choosing them one by one,u can't
include all of them as they are of order O(n^2).
so removing repetitions is the only option i think..On Wed, Dec 1, 2010 at 11:37 PM, fenghuang <fenghaungyuyi...@gmail.com>wrote: > @anoop > in fact, it always work even if there are repeated elements, because they > don't change the decision. > in detail, assume ii, ,jj, kk is one of the answers, then > a[ii]+a[jj]=a[kk]. since the array is sorted, so a[ii-1]+a[jj] <= a[kk] and > a[ii] + a[jj+1] >= a[kk]. > so when you try the pair of 'ii-1' and 'jj', the next step must be > calculate a[ii] + a[jj] as long as a[ii-1]+a[jj] is not equal to a[kk]. the > same to the pair 'ii' and 'jj+1'. > the algorithm is correct and in the whole procedure, repeated elements > don't affect the decision. > I'm sorry for my poor English. > Thank You! > > On Thu, Dec 2, 2010 at 12:14 PM, anoop chaurasiya < > anoopchaurasi...@gmail.com> wrote: > >> sorry for the interruption,we can make it work even if the elements are >> repeated, by removing the duplicacy in linear time(as the array is already >> sorted) and taking a count of no. of duplicates in the seperate array. >> >> On Wed, Dec 1, 2010 at 9:37 PM, Senthilnathan Maadasamy < >> senthilnathan.maadas...@gmail.com> wrote: >> >>> A small correction to the algorithm above. In Step 3, instead of finding >>> *any* pair (a,b) such that a+b = x we need to find *all* such pairs. >>> However the complexity remains the same. >>> >>> -- >>> You received this message because you are subscribed to the Google Groups >>> "Algorithm Geeks" group. >>> To post to this group, send email to algoge...@googlegroups.com. >>> To unsubscribe from this group, send email to >>> algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups.com> >>> . >>> For more options, visit this group at >>> http://groups.google.com/group/algogeeks?hl=en. >>> >> >> >> >> -- >> Anoop Chaurasiya >> CSE (2008-2012) >> NIT DGP >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to algoge...@googlegroups.com. >> To unsubscribe from this group, send email to >> algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@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 algoge...@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups.com> > . > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- Anoop Chaurasiya CSE (2008-2012) NIT DGP -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@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.
