I'm currently doing an exercise that asks to prove that Standard-Heapsort requires at fewest $\frac{1}{8} n \log(n) - O(n)$ comparisons, in its best case.
In its average case, Heapsort only requires $2n\log(n) - O(n)$ comparisons, although I don't know how to prove the best case scenario.
Can someone possibly hint me where to start with my calculation, I have no idea how to prove fewest number of comparisons, always been either worst case or most comparisons up until now.
