I am assuming you are talking about a binary heap, here is a simple case that shows the best case behavior.
Assume a binary heap of all identical elements.
Deletion from the binary heap is first switching the head with the last child, removing this child, and then making adjustments to ensure it is still a heap.
But in our case, after switching the head with the last element we still have root.value >= root.left.value && root.value >= root.right.value
, so we are done.
For this case, the number of operations is constant (regardless of the heap's size), so we can conclude this case is O(1)
, and since the best case cannot be better than O(1)
(there is no algorithm that runs better than that, in terms of asymptotic notation), we can conclude that the best case is O(1)
(tight bound).