find cumulative sum of each column. now for each arr[x][y] = sum of arr[i=0 to x] [j] ;
a1 b1 c1 d1 a2 b2 c2 d2 a3 b3 c3 d3 a4 d4 c4 d4 now we have reduced this problem to find max-subarray . which can be efficiently calculated using kadane's algo for each row. NOTE: now suppose if row = 0 does not participate in calculating max sum matrix so u need to subtract a1 from a2,a3,a4 .... similarly for other element in row 1,2,3. now updated matrix considered i.e from (row =1 to row =3 ). for this updated matrix, each[x][y] is the sum of arr[i=1 to x] [j]. similarly do for other elements. On Mon, Jan 16, 2012 at 6:55 AM, Ashish Goel <ashg...@gmail.com> wrote: > given a m*n matrix, find the subset rectangle with max sum (any other > rectangle taken would have lesser sum) > Best Regards > Ashish Goel > "Think positive and find fuel in failure" > +919985813081 > +919966006652 > > -- > 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.
