O(n^3) accepted fairly easy

I coded O(n^3) that was accepted fairly easy... But I know there is O(n^2) algo but I don't know how to implement it.. Any Ideas?

Re: O(n^3) accepted fairly easy

Posted by

Madhav 15 Jun 2008 15:04

Yaa i have the idea of O(n^2).First you can can calculate

the slopes of all lines in O(n^2).Create a class which contains the 2 points and the slope of the line joining two points.Note that there will be n(n+1)/2 nodes and define operator == as: slopes are equal and one point must be common to both the lines.

Re: O(n^3) accepted fairly easy

Posted by

majorro 13 Oct 2020 05:00

I have tl with long double and wa with double on that O(n^2) solution

*Edited by author 13.10.2020 05:01*