We can clearly see the pattern here,i.e.,everyf(k)insertions require O(lgr+f(k)) operations,then followed by a single insertion that requires O((r −k) + f(r)) operations. From this observation, we can approximate that the amortized cost is O(lgr+f(k)+(r−k+f(r))/). So with a large enough , the insertion cost in the worst case sceanrio is close to amortized O(lgr+f(k)) ≤ O(lgr+f(r)) operations, instead of O(r + f(r)) from Lemma 5, which is clearly an improvement.