i can improve a bit..
my logic is that i will take the a[n/2][n/2] element (middle of the diagonal element) if (num_2_b_search> a[n/2][n/2]) square sub matrix from a[0][0] till a[n/2][n/2] can be left out.. i mean divide the whole matrix in 4 half A B C D in my above case A is chopped off .. left alone is B C D we can now search recursively.. t(n)= 3 (t/4)+ constant t(n)= O(n^(log 3 base 4)) which is better than linear.. --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
