← Two Sigma Interview Insights
The core logic clicked pretty fast: just do ceiling division of the reading by the limit to get how many servers you need, then take the max with the current count.
First, clarify the problem: each server has a fixed capacity, and after each throughput reading, we need to ensure total capacity >= current throughput. We can maintain a running count of servers and add the minimum number needed to cover any deficit. The key is to compute the required servers as ceil(throughput / capacity) and take the maximum with the current count, since servers are never removed.
Pro tip: Mention that servers are never removed, so the count is non-decreasing. This simplifies the solution to a single pass with O(n) time and O(1) extra space, which is optimal.
Confirm that each server has a fixed throughput capacity, servers are never removed, and we need the minimum number of servers after each reading. Ask about input size and whether throughput readings are cumulative or per-interval.
At any step, total capacity = server_count * capacity_per_server. We need total capacity >= current throughput reading. So the required servers = ceil(throughput / capacity_per_server).
Initialize server_count = 0. For each throughput reading, compute required = ceil(reading / capacity). If required > server_count, set server_count = required. Append server_count to the result list.
Time complexity is O(n) for n readings, space O(1) extra (or O(n) for output). Handle edge cases: zero throughput, capacity zero (invalid), and large numbers (use integer arithmetic to avoid floating-point errors).
If servers could be removed, we might need a more complex data structure. If capacity varies per server, we'd need a different approach. Mention that the greedy approach works because capacity is additive and servers are identical.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.