Interestingly, they're doing unnecessary work here. In this case, the algorithm has already checked 32, so there's no need to check again. In fact, if you're c

Interestingly, they're doing unnecessary work here. In this case, the algorithm has already checked 32, so there's no need to check again. In fact, if you're checking N, it's guaranteed that N/2 has already been checked, and the next best step should be (3/4)N; in this case 48.