Slideworld wants to let the customers begin their trip at any landing b and end at any other landing e $>$ b$.$ Putting in all the $$n$$ slides in the park is going to break the bank. So you have been asked to determine a $2$ set of slides so that any customer can get from any landing b to any landing e $>$ b using at most two slides. That is$,$ it should be possible to get from any b to any e $>$ b either by taking a direct slide $($b $->$ e$)$ or by taking two slides $($b $->$ m$)$ and $($m $->$ e$).$ To be clear, a solution of your algorithm is a set of slides. Using divide$-$and$-$conquer, we will find a solution that uses only $\backslash$Theta $($n log n$)$ slides while ensuring that any customer can get from any landing b to any landing e $>$ b using at most two slides.