Yes, thats right. I think we can do the following also : Lets us assume rows are sorted in increasing order.
start from first row say i. Traverse the array from the end of the row towards the beginning till 0 occurs say at position j. now proceed to the next row i+1, check the value at i+1,j if it is 0, go to next row i+2,j else if its 1, then go to left till 0 occurs and store that index of 0 and follow to the next row. In the worst case, it will be O(n^2), but in general its a good approach i guess. what do u say guys ? Average Case O(m+n) ? Sanju :) On Sat, Aug 20, 2011 at 2:47 AM, shady <sinv...@gmail.com> wrote: > binary search on every row which will give solution in O(m*(logn)) > > On Sat, Aug 20, 2011 at 3:13 PM, Sanjay Rajpal <srn...@gmail.com> wrote: > >> Sorry I forgot to mention that. >> >> Sanju >> :) >> >> -- >> 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. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
