Shouldn’t this problem statement be to determine the critical height in the best-case rather than the worst-case ? As per my understanding, “worst-case” refers to the scenario when the case(s) always break, while in the “best-case” scenario, the case(s) never break and the critical height would be determined by the actual number of drops allowed.
24.29 Clarification to determine critical height problem
The worst-case here refers to the situation that no matter what the case critical height is. Your algorithm works always.
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