In the Bentley–Ottmann 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?