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Binary search on the answer, which I actually got to pretty fast.
Model the problem as finding the minimum time T such that the total number of customers served by all tellers within T minutes is at least m. Use binary search on T, where the feasibility check sums floor(T / serviceTimes[i]) for all tellers. Then prove correctness, analyze complexity, and implement with attention to edge cases.
Pro tip: Mention that the answer can be found by binary searching on the time, and that the upper bound can be set to min(serviceTimes) * m to avoid overflow and ensure efficiency. Also, discuss how to handle very large m by using 64-bit integers and early termination in the feasibility check.
Clarify that we need the minimum time T for at least m customers. The search space is from 0 to an upper bound like min(serviceTimes) * m, since the fastest teller alone can serve m customers in that time.
For a candidate T, compute the total customers served as the sum of floor(T / serviceTimes[i]) for all tellers. If this sum is >= m, T is feasible.
Binary search on T between low=0 and high=min(serviceTimes)*m. While low < high, compute mid, check feasibility, and adjust bounds accordingly. Return low as the answer.
Argue that the feasibility function is monotonic: if T works, any larger T also works. Thus binary search finds the minimum. Complexity: O(k log(min(serviceTimes)*m)) time, O(1) space.
Write code with 64-bit integers to avoid overflow. Handle cases like m=0 (return 0), very large m (ensure high bound doesn't overflow), slow tellers (serviceTimes large), and identical rates (sum simplifies to k * floor(T / rate)).
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