Logrithmic order can be solved... On Mon, Jul 18, 2011 at 11:49 PM, ankit sambyal <[email protected]>wrote:
> A more efficient approach : > Suppose the array is M*N > > i=j=0; > if(a[i][j] == x) > return; > mid1=(i+M-1)/2; > mid2=(j+N-1)/2; > > if(a[i][mid1]==x || a[mid2][j])==x > if(abs(a[i][mid1] - x) < abs(a[mid2][j]) - x) > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > 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 [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
