In the BentleyOttmann algorithm, Regarding :Find the segments r and t that are immediately below and above s in T  (if they exist) and if their crossing forms a potential future event  in the event queue, remove it.That means, every insertion of a segmant may lead to an intersection event deletion. I am thinking what makes it stay within the said time complexity. Is it valid to say that it is because i have up to n such insertions that would lead to up to n such deletions, which causes O(nlogn) operations at most hence the time complexity is not effected?
BentleyOttmann algorithm time complexity issue
computational geometry