ya rite.... On Wed, Mar 4, 2009 at 5:31 PM, Prunthaban <pruntha...@gmail.com> wrote:
> > @sharad - Your solution is O(n) Miroslav's solution is O(logn). > And what are you doing with the 'j' variable in your solution? Why not > simply use a[i] == k then c = i. That is what eventually your solution > is doing and it is O(n). > > > > On Mar 4, 4:32 pm, sharad kumar <aryansmit3...@gmail.com> wrote: > > pls tell wats difference btwn increasing and non decresing sorted > > array.question clearly tells n sorted array....... > > > > On Wed, Mar 4, 2009 at 4:55 PM, Miroslav Balaz <gpsla...@googlemail.com > >wrote: > > > > > of course, you cannot assume that array is in ascending order, it is in > > > non-decreasing order however not in ascending > > > and you should swap order here :(a[mid+1] > key||mid==high) > > > > > 2009/3/4 Kapil <navka...@gmail.com> > > > > >> just fixing a bug > > > > >> what if you write bin_search as this > > >> //assumption array is in ascending order > > >> binsearch(high, low, key, a) > > >> begin > > >> if low > high > > >> return -1 > > >> mid = (high+low)/2 > > >> if a[mid] = key And (a[mid+1] > key||mid==high) > > >> return mid > > >> if a[mid] <= key > > >> low = mid+1 > > >> else > > >> high = mid - 1 > > >> return binsearch(high,low,key,a) > > >> end > > > > >> On Mar 4, 3:46 pm, Kapil <navka...@gmail.com> wrote: > > >> > what if you write bin_search as this > > >> > //assumption array is in ascending order > > >> > binsearch(high, low, key, a) > > >> > begin > > >> > if low > high > > >> > return -1 > > >> > mid = (high+low)/2 > > >> > if a[mid] = key And a[mid+1] > key > > >> > return mid > > >> > if a[mid] <= key > > >> > low = mid+1 > > >> > else > > >> > high = mid - 1 > > >> > return binsearch(high,low,key,a) > > >> > end > > > --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
