Optiver Software Engineer Interview Questions

optiver fulltime software engineer technical phone interview experience

非leetcode. 让实现2个function来买和卖股票，要保证多线程下股票不能超卖（用lock）。我做出来了，但因为背景方向不match没进入下一轮。 Optiver只收junior SWE, 面试我的senior SWE/manager只是BS+5 yoe, 还没我资深(staff-level), 被猎头忽悠了来面。我以后应该不会面金融公司SWE了，要浪费几个小时做OA, 工资也不比硅谷AI岗高，要去就去quant/AI specialist.

System Design Prep Resources (Optiver)

Hey, I recently passed the behavioral phone call with Optiver and am now preparing for the technical round. While I expect the focus to be on DSA, I want to prep a little system design too. I have not

Optiver Online Test: Coding Challenge on News Subscription Management

The OA (Online Response Assessment) for Optimizer has three parts. Here's a sample coding question from the first part. A friend of mine also applied and encountered a question similar to the hot air

Optiver - SDE OA

A satellite network is experimenting with a distributed communication protocol based on satellite connections. During system tests, the team wants to ensure that messages they sent are processed in the correct...

Optiver Behavioral Phone Call

Hey! I just received an email for a 25-minute behavioral phone call from Optiver after completing my OA last week. This is my first interview and I am not sure what to expect. I assume it will mainly

Incomplete question in Optive coding OA

I received my Optiver OA assessment link last week, which contains three sections (in this post, I just want to talk about the coding part). The coding section consists of two questions (mostly OOP) w

Optiver | OA | Truck Positions

Truck Positions A logistics company has a central software solution to track the position of their trucks. Different applications are interested in different trucks. To save bandwidth the company wants to...

Optiver | OA | Concert Tickets

The OA was for Software Engineer Internship 2024 and below is the second question from it. # Concert Tickets ## Problem Statement As a long time live-music fan you have kept a bucket...

Optiver OA [ Hard ] @leetcode_dafu and @cpcs

The OA was for Software Engineer Internship 2024 and below is the second question from it. Concert Tickets Problem Statement As a long time live-music fan you have kept a bucket list of...

#622 Design Circular Queue

LeetCode #622: Design Circular Queue. Difficulty: Medium. Topics: Array, Linked List, Design, Queue. Asked at Optiver in the last 6 months.

#1206 Design Skiplist

LeetCode #1206: Design Skiplist. Difficulty: Hard. Topics: Linked List, Design. Asked at Optiver in the last 6 months.

#146 LRU Cache

LeetCode #146: LRU Cache. Difficulty: Medium. Topics: Hash Table, Linked List, Design, Doubly-Linked List. Asked at Optiver in the last 6 months.

Optiver Graduate SWE 2025 OA Q1: Fast Modular Arithmetic Under Constraints

## Problem You are given two integers `a` and `b` and a prime modulus `p`. Compute `(a^b) mod p` efficiently. Then: given an array of `n` integers, compute the product of all elements modulo `p`, but skip any element that is divisible by `p`. ```cpp long long mod_exp(long long base, long long exp, long long mod); long long filtered_product(vector<long long>& arr, long long p); ``` **Example:** ``` mod_exp(2, 10, 1000000007) -> 1024 filtered_product([2, 3, 7, 14, 5], 7) skip 7 and 14 (divisible by 7) -> (2 * 3 * 5) % 7 = 30 % 7 = 2 ``` ## Follow-ups 1. What is the time complexity of fast exponentiation? How does it compare to naive exponentiation for b = 10^18? 2. By Fermat's little theorem, what is `a^(p-1) mod p` when p is prime and gcd(a,p)=1? How can this simplify computing modular inverses? 3. If `p` is not prime, does your approach still work? What changes? 4. Implement `mod_exp` iteratively (no recursion) to avoid stack overflow for large exponents.

Optiver Graduate SWE 2025 OA Q2: Order Book Simulation with Price-Time Priority

## Problem Simulate a simplified limit order book. You receive a stream of orders: each order is `{id, side, price, quantity}` where `side` is `"buy"` or `"sell"`. - Buy orders are matched against the lowest-priced sell order at or below the buy price. - Sell orders are matched against the highest-priced buy order at or above the sell price. - On price tie, earlier orders have priority (FIFO). - Unmatched remainder stays in the book. After processing all orders, return the remaining orders in the book grouped by side. ```python def process_orders(orders: list[dict]) -> dict: # returns {"buy": [...], "sell": [...]} pass ``` **Example:** ``` Input: [{id:1, side:"sell", price:100, qty:5}, {id:2, side:"buy", price:102, qty:3}, {id:3, side:"buy", price:99, qty:2}] Order 2 matches order 1 for qty=3 @ price=100. Sell order 1 has 2 remaining. Order 3 (buy @99) cannot match sell @100. Output: {"buy": [{id:3,...}], "sell": [{id:1, qty:2,...}]} ``` ## Follow-ups 1. Which data structure gives you O(log n) insert and O(1) best-price lookup for each side? 2. How would you handle partial fills and order cancellations? 3. At 1 million orders/second, where are the bottlenecks in your implementation? 4. How would you persist the order book state across system restarts?