7 ms·
Even if the lightbulbs are breakable I think you would need a maximum of (2√n)-1 drops - i.e. go up in increments of √n then when it breaks work forward from th
by protothomas 16y ago
Even if the lightbulbs are breakable I think you would need a maximum of (2√n)-1 drops - i.e. go up in increments of √n then when it breaks work forward from the previous drop.
- Stormbringer 16y agoThat must be where they get the 20 figure from. My way has a shorter worst case even than that though, 14 vs 19