面试题库

30 天从前端工程师到 Agent 工程师

D8 为什么 Gateway/Worker 分离；Postgres 表设计（sessions/runs/messages）+ Drizzle

• 消息总线是至少一次投递，同一条消息被重复投递时，怎么保证不会产生两条 run？With at-least-once delivery, how do you guarantee a redelivered message does not create two runs?

  国内高频海外高频深入#idempotency#database#reliability

  分析过程 · 先想清楚再作答

  1. 这题在考幂等的落点在哪一层。凡是答「在代码里先查一下有没有，没有再插入」的，基本当场结束——因为那正是这题想筛掉的答案。
  2. 先把前提摊开：重复不是意外。总线是至少一次语义、客户端会超时重发、用户会手抖双击，同一句话到达两次是必然事件。所以设计目标不是「避免重复到达」，而是「重复到达时结果相同」。
  3. 然后给推导：幂等需要一个由请求内容决定的键。随机 UUID 每次都不同，等于没有幂等；正确取法是把会话 id、客户端消息 id、消息内容拼起来做哈希，客户端没有消息 id 时退用内容加一个粗粒度时间窗。
  4. 结论落在存储层：在 runs 表的这一列上加唯一约束，插入写成「冲突就什么都不做」，返回零行时回查那条已有的 run，把同一个 runId 返回给用户。两次请求、一条 run、一个 runId。
  5. 解释为什么「先查后插」不行，这是本题的分水岭：两个 Gateway 实例可以同时查、同时发现没有、同时插入，这两步之间有一个应用层拦不住的时间窗；它窄到压测复现不出来，上线后每天漏几条。**幂等的最终裁判必须是数据库的唯一约束**，应用层的判断只是为了少一次插入尝试。
  6. 可以预期的追问：那消费侧的重复执行呢？答：唯一约束保证了只有一条 run，但 Worker 可能重复拿到同一条 run，所以状态迁移也要带条件更新（只有当前状态是 pending 时才能改成 running），并且用一个显式的迁移白名单挡住「已完成的 run 被推回运行中」这种会覆盖用户已收到回复的情况。

  How to reason about it · think before answering

  1. This question is about which layer idempotency lives in. Anyone who answers 'check whether it exists, then insert' has usually just failed it — that is exactly the answer being screened out.
  2. State the premise: duplicates are not accidents. The bus is at-least-once, clients retry on timeout, users double-click. The same message arriving twice is certain, so the goal is not to prevent duplicates but to make duplicates produce the same result.
  3. Then derive the key: idempotency needs a key derived from request content. A random UUID differs every time and buys nothing; hash the session id, the client message id and the message body together, falling back to content plus a coarse time bucket when the client has no id.
  4. Land it in storage: put a unique constraint on that column in the runs table, write the insert as on-conflict-do-nothing, and when it returns zero rows read back the existing run and return the same run id. Two requests, one run, one id.
  5. Explain why check-then-insert fails, which is the whole point: two gateway instances can query, both see nothing, and both insert. The window between the two statements cannot be closed in application code, it is too narrow to reproduce under load tests, and it leaks a few bad rows every day in production. The database's unique constraint has to be the final arbiter; the application-level check only saves a wasted insert.
  6. Expect the follow-up: what about duplicate execution on the consumer side? The unique constraint gives you one run, but a worker can still receive it twice, so status changes need conditional updates (move to running only if the current status is pending) plus an explicit transition whitelist that blocks a finished run from being pushed back to running and overwriting a reply the user already saw.

  答题要点

  • 重复投递是必然事件，设计目标是「重复到达时结果相同」，不是「避免重复」
  • 幂等键必须由请求内容决定：会话 id 加客户端消息 id 加内容做哈希，随机 UUID 等于没有幂等
  • 在 runs 的幂等键列上建唯一约束，插入用「冲突就什么都不做」，零行时回查已有 run 返回同一个 runId
  • 先查后插在并发下必然出双份，两条语句之间的时间窗应用层拦不住，幂等的最终裁判是数据库唯一约束
  • 消费侧还要用条件更新加状态迁移白名单，避免同一条 run 被重复执行或把已完成的回复覆盖掉

  Key points

  • Redelivery is certain, so the goal is identical outcomes on duplicates, not preventing duplicates
  • The idempotency key must be derived from request content — session id plus client message id plus body, hashed; a random UUID buys nothing
  • Put a unique constraint on that column, insert with on-conflict-do-nothing, and read back the existing run when zero rows return
  • Check-then-insert races under concurrency; the window between the statements cannot be closed in application code, so the unique constraint must be the final arbiter
  • On the consumer side add conditional status updates and a transition whitelist so a finished run is never re-run or overwritten

D9 Redis Streams 消息总线：XADD/XREADGROUP/XACK/XAUTOCLAIM、consumer group、毒消息

• 什么是 at-least-once？既然消息会被重复投递，业务上到底要怎么保证幂等？What is at-least-once delivery, and given that messages get redelivered, how do you actually make the business side idempotent?

  国内高频海外高频深入#message-bus#idempotency#reliability

  分析过程 · 先想清楚再作答

  1. 这题是本章最容易被追到底的一道。绝大多数人能说出「至少一次，所以业务要幂等」，然后就没有下文了——面试官等的恰恰是下文：幂等具体落在哪一行代码上。答不出具体落点，前半句就是背的。
  2. 先解释为什么消费不掉这个重复：写业务和销号是两个系统的两次写（比如 Postgres 加 Redis），处理完成到 XACK 之间必然存在一个可以崩溃的窗口，崩在那里消息就会被重投。这个窗口只能变小，不能消失，所以 exactly-once 不是总线给你的语义。
  3. 由此得到一句可以直接说出口的结论：exactly-once 是消费端幂等做出来的**效果**，不是中间件提供的**能力**。Kafka 的事务能在「读 Kafka 写 Kafka」的闭环里做到，一旦下游是数据库或第三方 API 就又退回至少一次。
  4. 然后给具体落点，两道闸门要分清各自挡什么：第一道是 runs 表 idempotency_key 上的唯一约束，配 insert on conflict do nothing，挡的是**客户端重复提交**——冲突时接入层直接返回已有的 runId，连总线都不投第二遍；第二道是 messages 表的 unique(run_id, seq)，同样 on conflict do nothing，挡的是**同一条总线消息被执行两遍**，就算两个消费者真的同时跑完，用户也只会看到一条回复。中间还可以加一道便宜的短路：捞到消息先看 run 是不是已经 done，是就直接补一个 XACK 走人——但那是省钱的优化，正确性靠的是那两个唯一约束。
  5. 接着讲最容易做错的一步：幂等键怎么取。它必须能从「同一个意图」稳定推出来。客户端每次重试都新生成一个 uuid 是最常见的错法，那每次都是新意图，唯一约束一次都命中不了，闸门形同虚设。正确做法是客户端生成一次、重试复用同一个值，服务端兜底可以用「会话 id 加消息内容哈希加秒级时间戳」。
  6. 可以预期的追问：不可逆的副作用怎么办，比如发一次退款？答：把外部调用也变成带幂等键的（大多数支付网关都支持 idempotency key 头），并且先在本地库里落一条「已发起」记录再调用，用同一个键去重；实在不支持的接口就只能靠本地状态机加人工对账，这时要主动说出「这类操作我会把它挪出重试路径」。

  How to reason about it · think before answering

  1. This is the question that gets probed hardest. Most candidates say 'at-least-once, so make the business idempotent' and stop — but the follow-up is exactly what matters: which line of code enforces it.
  2. Explain why the duplicate cannot be removed: committing the business write and acking are two writes to two systems (say Postgres and Redis), so there is always a crash window between finishing the work and XACK. The window can shrink but not disappear, which is why exactly-once is not something the bus gives you.
  3. That yields a sentence worth saying out loud: exactly-once is an effect produced by consumer-side idempotency, not a capability provided by the broker. Kafka transactions achieve it inside a read-Kafka-write-Kafka loop, but the moment the sink is a database or third-party API you are back to at-least-once.
  4. Now name the concrete guards and what each one blocks. First, a unique constraint on runs.idempotency_key with insert ... on conflict do nothing, which blocks duplicate submissions: on conflict the gateway returns the existing run id and never publishes a second bus message. Second, unique(run_id, seq) on the messages table, also on conflict do nothing, which blocks duplicate execution: even if two consumers finish the same run simultaneously the user sees one reply. A cheap short-circuit can sit in between — read the run first and just re-ack if it is already done — but that saves money; correctness comes from the two constraints.
  5. Then the step people get wrong: deriving the key. It must be reproducible from the same intent. Generating a fresh uuid on every retry is the classic mistake, because every retry becomes a new intent and the constraint never fires. The client should mint the key once and reuse it across retries; a server-side fallback can hash session id plus message body plus a second-resolution timestamp.
  6. Expect: what about irreversible side effects such as issuing a refund? Push the idempotency key into the external call (most payment gateways accept an idempotency key header), and record an 'initiated' row locally before calling so the same key deduplicates. For APIs with no such support, fall back to a local state machine plus reconciliation, and say plainly that you would move such operations off the automatic retry path.

  答题要点

  • at-least-once：消息至少被处理一次、可能多次，因为业务提交和 XACK 是两个系统的两次写，中间的崩溃窗口消不掉
  • exactly-once 是消费端幂等做出来的效果，不是中间件的能力；下游只要是数据库或第三方 API 就退回至少一次
  • 闸门一：runs.idempotency_key 唯一约束 + on conflict do nothing，挡客户端重复提交，冲突时不再投递总线消息
  • 闸门二：messages 表 unique(run_id, seq) + on conflict do nothing，挡同一条消息被执行两遍，用户只会看到一条回复
  • 幂等键必须从同一个意图稳定推导，客户端重试要复用同一个值；每次重试新生成 uuid 等于没有幂等
  • 不可逆副作用要把幂等键透传给外部接口，并先落一条本地记录再调用

  Key points

  • At-least-once means a message is processed one or more times, because the business commit and the XACK are two writes to two systems with an unavoidable crash window
  • Exactly-once is an effect of consumer-side idempotency, not a broker feature; any database or third-party sink puts you back at at-least-once
  • Guard one: a unique constraint on runs.idempotency_key with on conflict do nothing blocks duplicate submissions and skips publishing a second bus message
  • Guard two: unique(run_id, seq) on messages with on conflict do nothing blocks duplicate execution, so the user sees exactly one reply
  • The idempotency key must be derivable from the same intent and reused across retries; minting a new uuid per retry defeats the whole mechanism
  • For irreversible side effects, pass the idempotency key through to the external API and record an initiated row locally before calling

D10 分片与租约：userId 哈希→shard、SET NX + TTL + Lua 续约、同用户顺序、handoff

• 两个 worker 同时认为自己持有同一个 shard 的租约（脑裂）会造成什么后果，怎么规避？What happens when two workers both believe they hold the same shard lease (split brain), and how do you mitigate it?

  国内高频海外高频深入#split-brain#fencing-token#reliability

  分析过程 · 先想清楚再作答

  1. 这题的判分点非常明确：答案里有没有出现「单靠 Redis 租约做不到绝对互斥」。说「用了 SET NX 加 TTL 就安全了」的人，会被追问到答不上来。
  2. 先讲脑裂是怎么发生的，而且要举那个最常见的场景——不是进程崩溃，是持有者只卡了 5 秒：一次 full GC、宿主机 CPU 被邻居打满、容器被 cgroup 限流。它醒过来时内存里还写着「我持有 shard 68」，继续处理手上那条消息、继续写库，而 Redis 里的租约早已到期并被别人抢走。再补一层：Redis 主从复制是异步的，切主时可能丢掉最后几毫秒的写入，于是两个 worker 都能 SET NX 成功。
  3. 然后讲后果，而且要落到业务上而不是停在「数据不一致」：同一个用户的两条消息被两个进程并发处理，回复乱序、上下文错乱、messages 表的 unique(run_id, seq) 撞约束导致落库失败；最严重的是有副作用的工具被重排或重复执行——「取消订单」和「改配送日期」顺序反了，结果是取消了一个用户本来想留下的订单。
  4. 关键的认知转折：既然无法在 Redis 一侧排除这条时间线，正确的思路就不是「让脑裂不发生」，而是「让第二个人的写入落不了地」——把冲突的检测与拒绝推到真正产生副作用的那一层。
  5. 三条手段按性价比给出。一是 worker 自己的自杀规则：连续两次续约失败、或距上次成功续约超过 TTL 的三分之二，立刻停止处理并清空持有集合——最便宜，把「我以为我还持有」的窗口从无限压到两个续约周期。二是 fencing token：抢租约时从一个单调递增计数器取号（Redis 的 INCR）写进租约值，之后所有有副作用的操作都带上它，下游只接受不比见过的最大号小的写入，落到数据库上就是一句条件更新；醒过来的前任拿的是旧号，写入直接被拒。三是每次写之前重新校验租约，并把校验与写入放进同一段脚本或同一个事务——这只缩小窗口，不消除。
  6. 可预期的追问：fencing 的局限在哪？答「它需要下游配合」。数据库能做条件更新所以好使，但下游是第三方接口（发短信、扣款）时你没法让对方帮你比号，这时只能退回幂等键，把重复执行变成无害，而不是让它不发生。真要绝对互斥就得换到有共识协议的系统（etcd、ZooKeeper 的会话租约），代价是写入延迟和运维复杂度。

  How to reason about it · think before answering

  1. The scoring criterion here is explicit: does your answer contain the sentence 'a Redis lease alone cannot give absolute mutual exclusion'. Anyone who says SET NX plus a TTL makes it safe gets probed until they run out of answers.
  2. Start with how split brain arises, and use the common case: not a crash, but a holder that merely froze for five seconds — a full GC, a noisy neighbour saturating the host CPU, cgroup throttling. It wakes up still believing it holds shard 68, keeps processing the in-flight message and keeps writing, while the lease expired and was taken. Add the second layer: Redis replication is asynchronous, so a failover can lose the last few milliseconds of writes and let two workers both win SET NX.
  3. Then the consequences, expressed in business terms rather than 'inconsistent data': two messages from one user processed concurrently means out-of-order replies, a corrupted context window, and unique(run_id, seq) violations that silently drop a message. Worst is reordered or duplicated side effects — swap 'cancel the order' with 'move the delivery date' and you cancel an order the user wanted to keep.
  4. The key shift: since you cannot rule out that timeline on the Redis side, the goal is not to prevent split brain but to make the second writer's writes fail — push conflict detection and rejection down to the layer that actually causes side effects.
  5. Give three mitigations by value. First, a self-kill rule in the worker: after two consecutive renewal failures, or when the last success is older than two thirds of the TTL, stop processing and clear the held set — cheapest, and it bounds the 'I think I still hold it' window to two renewal periods. Second, fencing tokens: take a monotonically increasing number (Redis INCR) when acquiring, store it in the lease value, attach it to every side-effecting operation, and have the downstream accept only numbers not lower than the highest it has seen — in a database that is one conditional update. The revived predecessor carries a stale number and is rejected. Third, re-validate the lease immediately before each write inside the same script or transaction, which shrinks the window without closing it.
  6. Expect the follow-up: where does fencing break down? It needs downstream cooperation. Databases do conditional updates, but a third-party endpoint (SMS, payments) will not compare your token, so you fall back to idempotency keys that make duplicate execution harmless rather than impossible. True mutual exclusion means moving to a consensus-backed system such as etcd or ZooKeeper session leases, paying in write latency and operational complexity.

  答题要点

  • 单靠 Redis 租约做不到绝对互斥：持有者被冻结再醒来、以及主从异步复制丢写，这两条时间线排除不掉
  • 后果要落到业务：同用户回复乱序、上下文错乱、唯一约束冲突丢消息，最严重是有副作用的工具被重排或重复执行
  • 思路是「让第二个人的写入落不了地」，把冲突检测推到产生副作用的那一层，而不是指望脑裂不发生
  • 三条手段：worker 自杀规则（续约连续失败就放手）、fencing token（写入时带单调号做条件更新）、写前重新校验租约
  • fencing 需要下游配合；下游是第三方接口时只能退回幂等键，要绝对互斥就得换 etcd / ZooKeeper 这类有共识协议的系统

  Key points

  • A Redis lease alone cannot guarantee mutual exclusion: a frozen holder that revives, and asynchronous replication losing writes on failover, are both unavoidable
  • State consequences in business terms: out-of-order replies, corrupted context, unique-constraint violations dropping messages, and reordered or duplicated side effects
  • The goal is to make the second writer's writes fail — push conflict detection to the side-effecting layer instead of hoping split brain never happens
  • Three mitigations: a worker self-kill rule on repeated renewal failure, fencing tokens enforced as conditional updates, and re-validating the lease immediately before writing
  • Fencing needs downstream cooperation; against third-party endpoints fall back to idempotency keys, and true mutual exclusion means a consensus system like etcd or ZooKeeper

D11 run 状态机、输出流回传、按 runId 保序、SSE 等待者、30s 打断合并

• 流式接口的客户端断线重连后，怎么做到既不丢片段也不重复？After a streaming client reconnects, how do you deliver every missed chunk exactly once — no gaps, no duplicates?

  国内高频海外高频深入#sse#idempotency#reconnect

  分析过程 · 先想清楚再作答

  1. 「不丢」和「不重复」要分开答。只答一半的人很多：说了从库里补发（不丢），却没说重叠部分怎么去重（不重复）。
  2. 推导链很短：客户端知道自己最后收到的编号 → 它重连时把这个编号带回来 → 服务端从下一号开始给 → 前提是编号连续不跳号。所以能不能重连，取决于当初有没有把序号设计成从 0 开始、连续、单调。序号一旦是时间戳或随机 id，这条链第一步就断了。
  3. 然后给三步实现和各自的坑：第一步换算，带回来的是「最后收到」的那一号，要加一，少加一重复一帧、多加一丢一个字，这是整段逻辑里唯一的算术也最常写错；第二步先从持久化里回放缺的部分，因为库里一定是全的；第三步再接上还在流动的那条流，两边必然重叠，靠「小于当前指针的一律丢弃」去重——幂等回放的全部秘密就是这一次比较。
  4. 这里要点出为什么必须双写：片段既进流也进库。只有流，重连时早期片段已被消费掉；只有库，就得轮询查库，首字延迟从几十毫秒涨到几百毫秒。代价是写放大，一次回答几百个片段就是几百行，生产里按批落库（每几十个片段或每两百毫秒一次）。
  5. 对齐一下协议细节：浏览器原生的事件源会自动把上次的编号放进重连请求头带回来；但大模型接口通常要用 POST，原生事件源只能发 GET，所以真实前端多是手写解析，重连时要自己把编号带上——这个细节能证明你真接过前端。
  6. 可以预期的追问：回放要保留多久？必须给两个边界——保留期（逐片段的行只对最近若干小时的执行保留，之后归档成一整条完整回复并删掉碎行）和回放上限（一次重连最多回放多少片段，超了就一次性发完整文本而不是逐字重演）。不定这两条，那张表会变成全库最大且 99% 的行写完十秒后再没人读。

  How to reason about it · think before answering

  1. Answer 'no gaps' and 'no duplicates' separately. Plenty of candidates cover only the first — they backfill from storage but never say how the overlap is deduplicated.
  2. The chain is short: the client knows the last id it received, it sends that id back on reconnect, the server resumes from the next one — and all of that requires contiguous, monotonic numbering. Whether resumption is possible at all was decided when you chose the sequence scheme; timestamps or random ids break the chain at step one.
  3. Then the three steps and their individual traps. Convert: the client reports the last id it *received*, so add one — off by minus one repeats a frame, off by plus one drops a character, and this is the only arithmetic in the whole flow and the most commonly wrong line. Replay: read the missing range from durable storage, which is always complete. Attach: resume the live stream, whose overlap with the replay is guaranteed, and drop anything below the cursor. That single comparison is all there is to idempotent replay.
  4. Explain why the dual write is mandatory: chunks go both to the stream and to the table. Stream only, and the early chunks are gone by reconnect time; table only, and you are polling the database, pushing time-to-first-token from tens to hundreds of milliseconds. The cost is write amplification — hundreds of rows per answer — so production batches the writes, every few dozen chunks or every couple hundred milliseconds.
  5. Get the protocol detail right: the browser's native event source replays the last id in a request header for you, but model endpoints generally need POST while that API only issues GET, so real frontends hand-roll the parser and must resend the id themselves. Mentioning this proves you have actually wired up the client side.
  6. Expect: how long do you keep replayable data? Give two bounds — a retention window (per-chunk rows only for runs from the last few hours, then collapsed into one complete message) and a replay cap (beyond N chunks, send the full text once instead of re-enacting it character by character). Without both, that table becomes the largest in the database while 99% of its rows are never read again after ten seconds.

  答题要点

  • 重连的前提是序号从 0 开始、连续、单调；序号是时间戳或随机 id 就无法续传
  • 客户端带回来的是「最后收到」的那一号，服务端要加一再开始，这是唯一的算术也最容易错
  • 先从库里回放缺的片段（库一定是全的），再接上还在流动的流，重叠部分靠「小于当前指针一律丢弃」去重
  • 片段必须双写：流服务当前挂着的连接，库服务等一下才回来的人；代价是写放大，生产里按批落库
  • 必须定保留期与回放上限：过期的执行归档成一整条完整回复，超长回放直接一次性发完整文本

  Key points

  • Resumption requires a contiguous, monotonic sequence starting at 0; timestamps or random ids make it impossible
  • The client reports its last received id, so the server resumes from that id plus one — the single most error-prone line
  • Replay the gap from durable storage first, then attach the live stream, discarding anything below the cursor to dedupe the overlap
  • Dual-write every chunk: the stream serves currently attached connections, the table serves clients that come back later; batch the writes in production
  • Set a retention window and a replay cap — archive old runs into one complete message and send full text instead of re-enacting long replays

D12 长期记忆：pgvector、embedding、chunking、memory_search 工具

• 把记忆检索包装成 memory_search 这样的工具时，参数设计上要注意什么？What matters when designing the parameters of a retrieval tool such as memory_search?

  国内高频海外高频深入#tool-design#security#long-term-memory

  分析过程 · 先想清楚再作答

  1. 这题看着是接口设计题，真正的区分度在安全。多数人会答 query 和 limit，答完就停；能不能说出「哪些参数绝对不能给模型」，决定了这题的分数。
  2. 先说 query：描述里要写清它不是用户原话，而是模型自己组织的检索语句，并给一个合法示例。用户问「我有什么忌口」，模型应该用「用户的食物忌口」去检索——这是从 D5 那条「格式类字段要给合法示例」延续下来的。
  3. 再说 limit：可选、默认 5、上限 10。上限的理由不是防呆，是上下文预算——一条记忆约 400 token，10 条就是 4000 token 进请求。模型传 50 时按工具协议回一条可读错误让它改，而不是静默截断成 10，否则模型永远不知道自己传错了。
  4. 然后是关键的一条：**绝不给 user_id 这类身份参数**。用户身份只能来自会话上下文。做成参数等于把「查谁的记忆」交给一段概率生成的文本，配上一句提示词注入就是现成的越权读取漏洞。一句话收尾：提示词管意图，代码管权限。
  5. 返回格式同样要说：空结果必须显式返回一句「没有找到相关记忆，请不要凭空推测」，返回空串模型会当成没有约束然后自己编；把相似度分数一起返回，模型才能区分「你说过」和「我印象里你好像提过」；设一条相似度下限，宁可不返回也不要拿噪声污染上下文。
  6. 可以预期的追问：检索不到的时候该怎么办？答案是分两层——工具层如实返回空并禁止推测，提示词层要求模型转而向用户确认，而不是把「没检索到」当成「用户没有偏好」。

  How to reason about it · think before answering

  1. It looks like an API design question; the discriminating part is security. Most candidates name query and limit and stop. Saying which parameters must never be exposed to the model is what earns the point.
  2. On query: the description must state that it is a retrieval phrase the model composes, not the user's literal words, and give a concrete example. Asked 'what are my dietary restrictions', the model should search for 'the user's food allergies and restrictions'.
  3. On limit: optional, default 5, capped at 10. The cap is a context budget, not idiot-proofing — a memory is roughly 400 tokens, so ten of them put 4000 tokens into the request. When the model asks for 50, return a readable validation error naming the field, the valid range and an example, rather than silently clamping, or it never learns it was wrong.
  4. The critical rule: never expose an identity parameter such as user_id. Identity comes from the session. Making it a parameter hands 'whose memories to read' to probabilistically generated text, and one prompt injection turns it into a privilege-escalation read. Prompts govern intent; code governs permission.
  5. Cover the response shape too: an empty result must say so explicitly and forbid guessing, because an empty string reads to the model as 'no constraints' and invites fabrication; return similarity scores so the model can distinguish a firm memory from a vague one; and set a minimum score, since no result beats a noisy one.
  6. Expect: what should happen on a miss? Two layers — the tool returns empty honestly and forbids speculation, and the prompt instructs the model to ask the user instead of treating 'not found' as 'no preference'.

  答题要点

  • query 必填，描述里说明它是模型组织的检索语句而非用户原话，并给一个合法示例
  • limit 可选、默认 5、上限 10，上限的依据是上下文预算；超限按工具协议回可读错误让模型改，不要静默截断
  • 绝不把 user_id 这类身份参数交给模型，身份只能来自会话——否则一句提示词注入就是越权读取
  • 空结果要显式说「没找到，请不要凭空推测」，返回空串模型会自己编
  • 返回相似度分数并设下限，宁可不返回也不要用低相关记忆污染上下文

  Key points

  • query is required; document it as a model-composed retrieval phrase, not the user's literal words, with an example
  • limit is optional, defaults to 5 and caps at 10 on context-budget grounds; over the cap, return a readable validation error instead of silently clamping
  • Never expose user_id or any identity parameter — identity comes from the session, or prompt injection becomes a privilege-escalation read
  • An empty result must say so explicitly and forbid speculation, or the model fabricates
  • Return similarity scores and enforce a minimum, since no result beats a noisy one

D13 cron 调度（中心调度→stream 投递）+ 成本计量（token→USD 台账、usage report）

• 怎么保证一个 cron 任务不会被重复投递或重复执行？幂等键应该怎么构造？How do you keep a cron job from being published or executed twice, and how should the idempotency key be built?

  国内高频海外高频深入#idempotency#scheduling#distributed-systems

  分析过程 · 先想清楚再作答

  1. 题眼在「投递」和「执行」是两件事。很多人只答一半：要么只说消费组保证一条消息一个消费者（那只挡住了执行侧的重复），要么只说加锁（那只挡住了投递侧，还挡不干净）。完整答案要说清两侧各自的重复来源，以及一个能同时兜住的兜底。
  2. 先拆重复的来源：投递侧的重复来自多个调度器实例、调度器重启后的补发重放、以及消息总线本身的至少一次语义；执行侧的重复来自 Worker 处理到一半崩溃后消息被 XAUTOCLAIM 转交给别人。这两类重复用不同手段挡效率完全不同。
  3. 再给核心结论：不要用分布式锁去做互斥，用数据层的唯一约束做去重。原因是锁只能提供「大概率互斥」——租约到期而前任进程其实只是 GC 卡住的那一瞬间，两个调度器都会认为自己持有，各发一次；而唯一约束是在最终落库那一步判断的，无论上游发了几次，任务表里只会多一行。能在唯一约束上解决的问题，不要升级成分布式协调问题。
  4. 然后回答幂等键怎么构造，这是最容易翻车的一步：键必须是「任务 id 加计划触发的那一分钟」，绝不能用当前时刻。两个调度器实例的时钟不可能对齐到毫秒，一个在 09:00:00.120 醒来、另一个在 09:00:00.480 醒来，用 now 算出来的键不一样，去重完全失效。把秒和毫秒截掉之后，无论谁在这一分钟里的哪一刻醒来，算出的键都是同一个字符串。落到代码上就是 insert 加 on conflict do nothing，冲突说明已经有人建过这次执行，直接 ack 掉不执行。
  5. 补一句作用范围：这套只保证「同一个触发点只产生一次执行」，不保证「执行内部的副作用只发生一次」。如果这次执行要发短信、要扣款，那些副作用还得各自带自己的幂等键，因为 Worker 可能在发完短信之后、写完状态之前崩掉。这一层区分是加分项。
  6. 可以预期的追问：那漏发怎么办？答宁可多发不可少发——调度器启动时回看最近 N 分钟逐分钟重放，重复投递被幂等键吃掉。at-least-once 加幂等是分布式系统里最省心的一组搭配，反过来先追求 exactly-once 再补幂等，通常两头都做不好。

  How to reason about it · think before answering

  1. The hinge is that publishing and executing are two separate problems. Most candidates answer half: either only the consumer group (which stops duplicate execution) or only a lock (which stops duplicate publishing, and imperfectly). A complete answer names the duplicate sources on both sides plus one backstop that covers both.
  2. Enumerate the sources: duplicate publishes come from multiple scheduler instances, from replay after a scheduler restart, and from the bus's own at-least-once semantics. Duplicate executions come from a worker crashing mid-processing and the message being reclaimed by another consumer. The two need different treatment.
  3. State the core conclusion: do not reach for a distributed lock, use a uniqueness constraint in the data layer. A lease only gives you probable mutual exclusion — in the instant when the TTL expires while the previous holder is merely stuck in GC, both schedulers believe they hold it and both publish. A uniqueness constraint is evaluated at the final insert, so no matter how many times upstream published, the table gains exactly one row. Do not escalate a problem solvable by a constraint into a distributed coordination problem.
  4. Then the key construction, which is where people fail: the key must be the task id plus the scheduled minute, never the current instant. Two scheduler clocks never align to the millisecond; one wakes at 09:00:00.120 and the other at 09:00:00.480, so keys built from now differ and dedup collapses. Truncate seconds and milliseconds and every instance computes the same string for that minute. In code this is an insert with on conflict do nothing; a conflict means the execution already exists, so ack the message and skip.
  5. Scope it honestly: this guarantees one execution per trigger point, not that side effects inside the execution happen once. If the run sends an SMS or charges a card, those side effects need their own idempotency keys, because the worker can crash after sending and before writing status. Making that distinction earns points.
  6. Expect: what about missed triggers? Prefer over-publishing to under-publishing — replay the last N minutes at startup and let the idempotency key absorb duplicates. At-least-once plus idempotency is the easiest combination in distributed systems; chasing exactly-once first and adding idempotency later usually achieves neither.

  答题要点

  • 投递重复和执行重复是两件事：前者来自多调度器实例、重启重放和总线的至少一次语义，后者来自 Worker 崩溃后消息被转交
  • 用数据层唯一约束去重，不要用分布式锁互斥：租约过期而前任还活着的瞬间两个调度器都会各发一次，而唯一约束在落库那一步只放行一条
  • 幂等键必须是任务 id 加计划触发的那一分钟，不能用当前时刻——两个实例的醒来时刻永远不同，用 now 会让去重完全失效
  • 落到代码上是 insert 加 on conflict do nothing，冲突就直接 ack 不执行
  • 这只保证一个触发点一次执行，执行内部的发短信、扣款等副作用要各自带幂等键；宁可多发不可少发，靠 at-least-once 加幂等兜底

  Key points

  • Duplicate publishing and duplicate execution are separate: the former comes from multiple schedulers, restart replay and at-least-once delivery; the latter from a crashed worker's message being reclaimed
  • Dedupe with a database uniqueness constraint rather than a distributed lock: when a lease expires while the holder is only GC-stalled, both schedulers publish, whereas the constraint admits exactly one row
  • Build the key from the task id plus the scheduled minute, never the current instant — instances never wake at the same millisecond, so a now-based key defeats dedup entirely
  • In code this is an insert with on conflict do nothing; on conflict, ack the message and skip execution
  • This guarantees one execution per trigger, not once-only side effects — SMS or payments inside the run need their own keys; prefer over-publishing and let at-least-once plus idempotency absorb it

D14 部署运维：compose 多 worker、心跳、健康检查、优雅停机、dev/prod 隔离；W2 复盘

• 系统设计：请设计一个 IM Agent 平台——用户在即时通讯软件里和一个 AI 助手对话，助手能调用工具、记住长期偏好、还能定时主动推送。要求支撑十万日活。System design: design an IM agent platform where users chat with an AI assistant inside a messaging app. The assistant calls tools, remembers long-term preferences, and proactively pushes scheduled messages. Target 100k daily active users.

  国内高频海外高频深入#system-design#distributed-systems#cost#operations

  分析过程 · 先想清楚再作答

  1. 先别画图。系统设计题最常见的死法是听完就开始画框，二十分钟后面试官发现你解的是另一道题。花三到五分钟问清四件事：一是流量形状（十万日活对应多少并发会话、峰谷比多少），二是延迟要求（首字节要多快，是否必须流式），三是工具的性质（只读查询还是有写操作和副作用），四是主动推送的合规边界（能不能在深夜推、每天上限几条）。这四个答案会实质改变架构，问它们本身就是分数。
  2. 然后给主干，一句话先定形状：**接入层无状态、消息总线解耦、Worker 有状态且按用户分片、状态全在数据库**。接着按数据流走一遍：IM 平台的 webhook 打到接入层，接入层只做鉴权、限流、落库、投递四件事，立刻返回 202；执行侧从总线取活、跑 Agent 循环、把输出片段回传；主动推送由一个中心调度器按时间投递进同一条总线。**关键论点是接入层耗时确定、执行层耗时不确定，把它们放在一个进程里意味着一次慢的模型调用会占住一个本该毫秒级返回的连接**——这是整道题的立论基础，要主动说出来。
  3. 再逐个模块给出选择和理由。存储：sessions / runs / messages 三张表，runs 单独存在是因为只有它能回答「这次到底跑完没有」，幂等靠 runs 上的唯一约束而不是先查后插。总线：Redis Streams 的消费组做分摊，语义是至少一次，恰好一次靠消费端幂等做出来；反复失败的消息投递三次后进死信流。顺序：消费组的分配单位是一条消息而业务要求的串行单位是一个用户，所以按 userId 哈希到固定数量分片，每个分片同一时刻只有一个 Worker 持有租约。记忆：pgvector 存 embedding，检索包成一个工具交给模型自己决定要不要查，且不给它身份参数——身份只能来自会话。
  4. 主动推送这一块要单独讲透，因为它是这道题区别于普通聊天服务的地方。中心调度器命中时间点后只投一条消息，执行侧照旧；幂等键锚在「计划触发的那一分钟」，所以调度器崩溃重启后回看重放不会重复推送。合规上要有时区、静默时段、每日上限三道闸，而且这三道闸必须在投递前判断而不是在推送时判断——否则你已经花了模型调用的钱才发现不该推。
  5. 然后主动给出容量和成本的数字感，这是高级候选人的分水岭。十万日活、人均十轮对话是一百万次模型调用；按输入输出各一千 token、每百万 token 输入 0.15 美元输出 0.60 美元估算，一天大约七百五十美元。这个数字立刻推出三件事必须做：token 用量要按调用记账并换算成美元（否则你无法定位是哪个用户或哪个功能在烧钱）、要有分层降级（超预算的用户切便宜模型而不是直接拒绝）、以及上下文长度是主要成本杠杆（所以要压缩历史、控制检索条数）。
  6. 最后收在可运维性上，也就是这一周的落点：多副本部署、心跳发现假死、就绪探针只查自己必需的依赖、优雅停机让发版不掐断对话、dev 与 prod 用键名前缀隔离。**每个机制都要配一句「它失效时会怎样」**——租约会脑裂所以要有自杀规则和护栏令牌、心跳会误判所以面板转红只告警不自动摘流量、停机会超时所以等待要有上限。说不出失效模式的机制，面试官会认为你只是读过。
  7. 可以预期的追问，按出现频率排：单点在哪（调度器无状态可重启，Redis 和 Postgres 靠托管服务的主备）；怎么灰度（新旧 Worker 同时在线，靠消息里的版本字段决定走哪套提示词）；用户在助手回复中途又发一句怎么办（三十秒内的改口合并进同一次执行，而不是并发开两个）；成本再降一半怎么做（缓存高频问答、压缩历史、把简单意图路由到小模型）。

  How to reason about it · think before answering

  1. Do not start drawing. The most common way to fail a design question is to hear the prompt and immediately sketch boxes, only for the interviewer to realise twenty minutes later that you solved a different problem. Spend three to five minutes on four questions: traffic shape (how many concurrent sessions does 100k DAU imply, and what is the peak-to-trough ratio), latency (how fast must first byte be, is streaming required), the nature of the tools (read-only lookups, or writes with side effects), and the compliance boundary on proactive pushes (may you push at night, what is the daily cap). All four change the architecture materially, so asking them is itself worth points.
  2. Then state the trunk in one sentence: stateless ingress, a message bus for decoupling, stateful workers sharded by user, all state in the database. Walk the data flow: the messaging platform's webhook hits ingress, which does only auth, rate limiting, persistence and publish, and returns 202 immediately; the execution side pulls work, runs the agent loop, and streams output fragments back; proactive pushes come from a central scheduler publishing onto the same bus. The load-bearing argument is that ingress latency is bounded while execution latency is not, so putting them in one process means one slow model call occupies a connection that should have returned in milliseconds — say this out loud, it is the premise of the whole answer.
  3. Then justify each module. Storage: sessions, runs and messages, with runs existing separately because only it can answer whether this attempt actually finished; idempotency comes from a unique constraint on runs, not from check-then-insert. Bus: Redis Streams consumer groups for fan-out, at-least-once semantics, with exactly-once manufactured by consumer-side idempotency, and messages that fail three times moved to a dead-letter stream. Ordering: the consumer group's unit of assignment is one message while the business requires serialisation per user, so hash userId into a fixed set of shards and let exactly one worker hold each shard's lease. Memory: embeddings in pgvector, retrieval wrapped as a tool the model chooses to call, with no identity parameter — identity only ever comes from the session.
  4. Treat proactive push as its own section, because it is what separates this from an ordinary chat service. The central scheduler publishes one message on a time match and the execution side is unchanged; the idempotency key is anchored to the scheduled minute, so replaying after a scheduler restart cannot double-send. For compliance you need timezone, quiet hours and a daily cap — and all three must be evaluated before publishing rather than at send time, or you have already paid for the model call before discovering you should not have pushed.
  5. Then volunteer capacity and cost numbers, which is what separates senior candidates. 100k DAU at ten turns each is a million model calls; at roughly a thousand tokens in and out, with input at $0.15 and output at $0.60 per million tokens, that is about $750 a day. That number immediately implies three requirements: meter token usage per call and convert to dollars (otherwise you cannot tell which user or feature is burning money), build tiered degradation (push over-budget users to a cheaper model rather than refusing them), and recognise that context length is the dominant cost lever (so compress history and cap retrieved items).
  6. Land on operability, which is this week's payoff: multiple replicas, heartbeats to surface zombies, readiness probes that only check their own hard dependencies, graceful shutdown so deploys do not cut conversations, and dev/prod isolation via key prefixes. Pair every mechanism with what happens when it fails — leases can split-brain so you need a self-fencing rule and fencing tokens, heartbeats produce false positives so a red dashboard alerts a human rather than auto-draining, shutdown can time out so the wait needs a ceiling. A mechanism without a stated failure mode reads as something you only read about.
  7. Expect, in rough order of frequency: where are the single points (the scheduler is stateless and restartable; Redis and Postgres rely on managed primary/replica); how do you roll out safely (old and new workers coexist and a version field in the message selects the prompt set); what if the user sends another message mid-reply (merge a change of mind within thirty seconds into the same execution rather than running two concurrently); and how would you halve the cost (cache frequent answers, compress history, route simple intents to a smaller model).

  答题要点

  • 先用三到五分钟问清四件事：流量形状、延迟要求、工具是否有副作用、主动推送的合规边界——它们会实质改变架构
  • 主干一句话：接入层无状态、消息总线解耦、Worker 有状态且按用户分片、状态全在数据库；立论是接入层耗时确定而执行层不确定
  • 存储 sessions / runs / messages 三张表，幂等靠 runs 上的唯一约束；总线用 Redis Streams 消费组，至少一次加消费端幂等，三次失败进死信
  • 顺序靠 userId 哈希分片加租约：消费组的分配单位是一条消息，而业务要求的串行单位是一个用户
  • 记忆用 pgvector 并包成工具交给模型自己决定是否检索，不给身份参数——身份只能来自会话
  • 主动推送由中心调度器投递，幂等键锚在计划触发的那一分钟；时区、静默时段、每日上限三道闸必须在投递前判断
  • 给出成本数字感：十万日活人均十轮约一百万次调用、一天约七百五十美元，由此推出计量记账、分层降级、压上下文三件事
  • 收在可运维性：多副本、心跳查假死、就绪探针只查自己的依赖、优雅停机、dev/prod 前缀隔离
  • 每个机制都配一句失效模式：租约会脑裂、心跳会误判、停机会超时——说不出失效模式等于只是读过

  Key points

  • Spend three to five minutes clarifying four things: traffic shape, latency targets, whether tools have side effects, and the compliance boundary on proactive pushes
  • State the trunk in one sentence: stateless ingress, bus for decoupling, stateful workers sharded by user, all state in the database — premised on bounded ingress latency versus unbounded execution latency
  • Storage is sessions/runs/messages with idempotency from a unique constraint on runs; the bus is Redis Streams consumer groups, at-least-once plus consumer idempotency, dead-lettering after three failures
  • Ordering comes from hashing userId into shards plus leases: the consumer group assigns per message while the business serialises per user
  • Memory is pgvector exposed as a tool the model may call, with no identity parameter — identity comes only from the session
  • Proactive push flows through a central scheduler with the idempotency key anchored to the scheduled minute; timezone, quiet hours and daily caps are enforced before publishing
  • Bring numbers: 100k DAU at ten turns is ~1M calls and ~$750/day, which implies metering, tiered degradation, and context length as the main cost lever
  • Land on operability: replicas, heartbeats for zombies, readiness probes scoped to own dependencies, graceful shutdown, dev/prod prefix isolation
  • Pair each mechanism with its failure mode — leases split-brain, heartbeats false-positive, shutdown times out; a mechanism without one reads as book knowledge