take two pointer p and q p=a[0] and q=a[n-1]; sum=p+q; if(sum>x) q--; if (sum<x) p++;
On Sun, Oct 10, 2010 at 6:54 PM, Rohit <[email protected]> wrote: > > > On Oct 10, 10:48 am, Shravan <[email protected]> wrote: > > http://ideone.com/D5W2y > > > > Good :). > What if array is unsorted. What will be the best solution in terms of > time complexity? > Better then (nlog(n)+n). > > Regards > Rohit > > -- > 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]<algogeeks%[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.
