I was pretty careless with cut and paste from Case 1. Let's try Case 2 again Start: 1 1 0 1 1 1 0 1 1
After first scan: 1 1 0 1 1 1 0 1 0 After second scan: 0 1 0 0 1 1 0 1 0 After third scan: 0 0 0 0 1 1 0 1 0 It looks like I need a 4th step. If the lower right element is zero, zero the last row and the last column. 0 0 0 0 1 0 0 1 0 The simplest case which requires this missing step is 1 1 1 1 1 1 1 1 0 which should produce 1 1 0 1 1 0 0 0 0 The four-step algorithm can be streamlined slightly as follows: Scan the array. If the i,j element is zero, set the last element of row i and the last element of column j to zero. Second, scan the last row, ignoring the last element, and zero every column that ends with a zero. Third, scan the last column, including the last element, and zero every row that ends with a zero. Finally, if the last element in the last row is zero, zero the last column. This touches every element at most 3 times; i.e., if the array has m rows and n columns, the algorithm is O(1) in space and O(m*n) in time. Dave On Nov 23, 2:20 am, "chandra kumar" <[EMAIL PROTECTED]> wrote: > I have some questions in case 2.... questions inlined.. > > Thanks and regards, > K.V.Chandra kumar. > > On 22/11/2007, Dave <[EMAIL PROTECTED]> wrote: > > > > > > > Case 1 > > Start: > > 0 1 1 > > 1 1 1 > > 1 1 1 > > > After first scan: > > 0 1 0 > > 1 1 1 > > 0 1 1 > > > After second scan: > > 0 1 0 > > 0 1 1 > > 0 1 1 > > > After third scan: > > 0 0 0 > > 0 1 1 > > 0 1 1 > > > Case 2 > > Start: > > 1 1 0 > > 1 1 1 > > 0 1 1 > > > After first scan: > > 0 1 0 how did we get 0 in the first row first column, we > > should only alter the last row and column, right? > > 1 1 1 > > 0 1 0 > > > After second scan: > > 0 0 0 how did we get these zero, only columns are to be > > changed, right? > > 0 1 1 > > 0 0 0 > > > After third scan: > > 0 0 0 > > 0 1 0 how did we get this as, ast element is ignored, right? > > 0 0 0 > > > Hope this helps. > > Dave > > > On Nov 22, 9:34 am, "chandra kumar" <[EMAIL PROTECTED]> > > wrote: > > > Hi Dave, > > > Can you explain your algo for these 2 cases... > > > > 0 1 1 1 1 0 > > > 1 1 1 1 1 1 > > > 1 1 1 0 1 1 > > > > Please explain me in steps cause we tried the same problem and can't > > get > > > it done for without using extra space. ( we used 1 extra space, if the > > > array can contain only 0 or 1 and no other number, otherwise we store > > some > > > other number indicating a yet to become zero and proceed, in the later > > case > > > we need no extra space ) > > > Thanks and Regards, > > > K.V.Chandra Kumar > > > > On 15/11/2007, Dave <[EMAIL PROTECTED]> wrote: > > > > > Scan the array. If the i,j element is zero, set the last element of > > > > row i and the last element of column j to zero. Then, scan the last > > > > row, ignoring the last element, and zero every column that ends with a > > > > zero. Similarly, scan the last column, ignoring the last element, and > > > > zero every row that ends with a zero. This touches every element at > > > > most 3 times; i.e., if the array has m rows and n columns, the > > > > algorithm is O(1) in space and O(m*n) in time. > > > > > Dave > > > > > On Nov 14, 1:27 pm, geekko <[EMAIL PROTECTED]> wrote: > > > > > Given an array of 0's and 1's whenever you encounter an 0 make > > > > > corresponding column and row elements 0. How could you do that > > > > > efficiently(minimum time and minimum space complexity)? This > > question > > > > > is taken from placementsindia.blogspot.com- Hide quoted text - > > > > - Show quoted text -- Hide quoted text - > > - Show quoted text - --~--~---------~--~----~------------~-------~--~----~ 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 [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---
