Optiver Interview Tips: VO Questions Answers & OA Interviews

Optiver Interview Aid

To become a Salesforce Engineer, your journey starts with a strong resume - recruiting systems and teams will focus on candidates who have hands-on experience with core Salesforce technologies like Apex and LWC, as well as basic technologies like Java and SQL. After your resume is approved, you'll likely have to face a round of online programming challenges (OA) or technical phone interviews, which are preliminary "feelers": OA tests your algorithmic and programming fundamentals, while phone interviews are conducted by engineers in person, who will talk to you about the projects you've worked on, your understanding of the platform's features, and how to use code to solve real business challenges. The key to this step is to prove that you have not only "used" Salesforce, but really understand its development logic.

If you make it to the next stage, it's really multiple rounds of "hardcore" interviews, often in intense live or virtual sessions, that will thoroughly evaluate you on multiple dimensions: there will definitely be a round focusing on core platform technical details such as Apex programming depth and Governor Limits optimization; another round will have you playing the role of an architect, discussing the design and integration of a large Salesforce system in depth; and another round will have you playing the role of an architect. One round will definitely focus on core platform technical details like Apex programming depth and Governor Limits optimization; another will have you play architect and discuss in-depth the design, integration, and scalability of a large Salesforce system. Of course, there's also the behavioral interview, in which the interviewer will judge whether you fit Salesforce's emphasis on a culture of innovation and collaboration through your experience of dealing with difficulties and working collaboratively. The process is notoriously difficult, and the goal is clear: to find top engineers who understand platform technology, but also have excellent design skills and good professionalism.

Interviews on behalf of major North American technology companies

OA Interview Questions Example Answers

title: Given a set of assets, each with expected return and risk, choose a subset to maximize the Sharpe ratio, given the constraints.

This is a typical portfolio optimization problem, which is very common in practical quantitative trading. Since the question asks to choose a subset of assets to maximize the Sharpe ratio and does not give specific constraints and a definition of risk (e.g., whether it is a covariance matrix or simply standard deviation), we assume that the asset risk is the standard deviation and that the constraints refer mainly to the number of assets or the total investment amount.

Problem abstraction and mathematical modeling

Sharpe Ratio ($S$) is defined as:

S = \frac{E(R_p) - R_f}{\sigma_p}

where $E(R_p)$ is the expected return of the portfolio, $\sigma_p$ is the risk (standard deviation) of the portfolio, and $R_f$ is the risk-free interest rate (in competitive programming, it is common to assume that $R_f = 0$ to simplify the problem). Thus, the objective is to maximize $\frac{E(R_p)}{\sigma_p}$. Suppose we have

N

individual assets, the decision variables are binary:

x_i \in \{0, 1\}

Indicates whether the asset is selected

i

The

Portfolio returns $E(R_p) = \sum_{i=1}^{N} x_i E(R_i)$ which $E(R_i)$ It's an asset. $i$ of the expected returns.
Portfolio risk $\sigma_p$ The calculation depends on the correlation between the assets.
- Stand-alone asset assumption (simplified version): If the assets are independent, the $\sigma_p = \sqrt{\sum_{i=1}^{N} x_i^2 \sigma_i^2} = \sqrt{\sum_{i=1}^{N} x_i \sigma_i^2}$ which $\sigma_i$ It's an asset. $i$ The standard deviation of the
- Relevant assets (true condition): $\sigma_p = \sqrt{\sum_{i=1}^{N} \sum_{j=1}^{N} x_i x_j \sigma_{ij}}$ which $\sigma_{ij}$ It's an asset. $i$ respond in singing $j$ The covariance of the

Since this is a subset selection problem, the decision variables are discrete (0 or 1), and the objective function is usually nonlinear (because of square roots and divisors), this is a Non-Linear Integer Non-Linear Programming (MINLP) problem, which is an NP-hard problem.

Greedy or backtracking based search

Faced with the problem of subset selection for NP-hard, especially in competitive programming or OA, it is often necessary to consider approximation algorithms, heuristic searches, or brute force searches with small constraints:

Less constrained (number of assets) $N$ Smaller, as in $N \le 20$ ): Brute force enumeration can be performed using Backtracking or State Compressed Dynamic Programming (Bitmask DP) for all $2^N$ subsets, calculate the Sharpe ratio for each subset and take the maximum value.
When constraints are large (number of assets) $N$ (Larger): Violent enumeration is not feasible. If correlations between assets are ignored (i.e., assumed to be independent) and the objective function becomes more tractable, a greedy algorithm or dynamic programming can be attempted. If correlations cannot be ignored, heuristic searches such as Simulated Annealing or Genetic Algorithm are needed, but these are usually out of scope in OA problems unless the problem is specifically simplified.

Assuming assets are independent and using backtracking search

To give an executable solution process, we adopt a simplification common in OA problems: assume that the assets are independent and that the number of assets

N

smaller, using a backtracking search (Depth-First Search, DFS):

Initialization: Maintains a global variable max_sharpe to record the maximum Sharpe ratio found.
DFS function definitions: DFS(index, current_return, current_variance)
- index: Serial number of the currently considered asset.
- current_return: The currently selected asset'sTotal expected returnThe
- current_variance: The current selection of the asset'stotal variance( $\sum \sigma_i^2$ ).
Recursive termination conditions:
- If index == N (all assets have been considered), the Sharpe ratio of the current portfolio is calculated: $S = \frac{current\_return}{\sqrt{current\_variance}}$ If $current\_variance = 0$, $S$ is set to $\infty$ (if $current\_return > 0$) or 0. Update max_sharpe = max(max_sharpe, S).
Recursive steps (for the first $index$ (an asset):
- No choice of assets $index$ :: Call DFS(index + 1, current_return, current_variance)
- Selection of assets $index$ ::
  - Update return: new_return = current_return + E(R_index)
  - Update variance: new_variance = current_variance + \sigma_{index}^2
  - Call DFS(index + 1, new_return, new_variance)

Binding treatment: If there is a limit (e.g., "select up to K assets"), you can add a counter parameter to the DFS function and recursively select assets by checking to see if more than

K

The

explicate

The final answer to the solution is the maximum max_sharpe value found during the backtracking search (DFS) and the subset of assets that correspond to the assets that produced that maximum Sharpe ratio. If the input data is: | Assets | Returns $E(R_i)$ | Risk $\sigma_i$ | $\sigma_i^2$ | | :-: | :-: | :-: | :-: | :-: | | A | 0.10 | 0.05 | 0.0025 | | B | 0.20 | 0.10 | 0.0100 | | C | 0.15 | 0.05 | 0.0025 | A | 0.10 | 0.05 | 0.0025 The result of the violent enumeration will be: | combination | $E(R_p)$ | $\sigma_p$ | $S$ | | :-: | :-: | :-: | :-: | :-: | | {B, C} | 0.35 | $\sqrt{0.0125} \approx 0.1118$ | $\approx 3.13$ | | ... | ... | ... | ... | ... | ... | ... | ... | ... With a complete DFS traversal, we are guaranteed to find the subset of assets that maximizes the Sharpe ratio. The core difficulty of this question lies in the subset selection (discrete decision making) and the nonlinear objective function (Sharpe ratio). If the question does not explicitly give a simplification (e.g., asset independence) and allows for investment ratios (weights)

w_i \in [0, 1]

), then it becomes a quadratic programming problem with continuous variables that can be solved using more efficient numerical optimization methods (e.g., Newton's method or interior point methods), which is another class of classical Markowitz optimization problems. However, for the

0/1

subset selection, backtracking search is the most straightforward and reliable method when the number of assets is small.

VO Interview questions

To process a real-time, continuously incoming stream of price data, you need to design an efficient data structure and algorithm for streaming prices over a length of

K

within a sliding window, the median of all data within the current window is calculated in real time. request::

Design an add(price) method so that the data structure can be updated when a new price data comes in.
Design a getMedian() method that returns the current median as quickly as possible. $K$ The median of the individual data points.
When new data comes in, if the window size is more than $K$ , the oldest data points need to be removed.

suppose that...

K

is a larger odd number.

Solution Ideas and Processes:

The difficulty of this question lies in two operations: adding/deleting data points (sliding window) and quickly querying the median. While traditional arrays or chained lists can be easily added and deleted, querying the median requires sorting, and the time complexity is

O(K \log K)

, which is too slow for real-time systems. A balanced binary search tree (BST) or a skiplist can do

O(\log K)

The insertion/deletion and

O(\log K)

queries, but the implementation is complex. The cleanest, most efficient, and most desired solution by the interviewer is to use the structure of **Two Heaps** on the top heap (Two Heaps)**.

Core data structure: top-to-top heap

We use two heaps to maintain the data within the window:

MaxHeap / Smaller Half: Stores the smaller half of the window. The top of the heap is the maximum value of the "smaller half".
Minimum Heap (MinHeap / Larger Half): Stores the larger half of the data in the window. The top of the heap is the smallest of these "larger halves".

By maintaining the following two key attributes, we can obtain the median in real time:

The size of MaxHeap is either equal to the size of MinHeap or one element more than it (because the $K$ (is odd, there is always a heap bigger).
All elements in MaxHeap are less than or equal to all elements in MinHeap.

In this state of equilibrium, the median is the top element of the MaxHeap.

Algorithm implementation steps

To handle sliding windows **add (

O(\log K)

) and deletion (

O(\log K)

For this operation, we need an additional data structure to keep track of the position of each element and to implement "deferred deletion" or "efficient lookup deletion" of the heap. HashMap** is chosen to keep track of the elements and their frequencies in the heap to support

O(\log K)

The deletion of the A. General overview of data structures:

MaxHeap: Store the smaller half of the data.
MinHeap: Stores the larger half of the data.
HashMap (counts): Record the number of occurrences of each price in the window, used to handle remove operations.
Deque (window): Maintains the order of elements within a window, and is used to identify the oldest element to remove.

B. add(price) operation:

Add new elements: Add price to the end of the window queue. Insert price into the appropriate heap (if price is less than the top of the MaxHeap heap, it goes into MaxHeap; otherwise it goes into MinHeap). Update counts.
Balance the pile: If MaxHeap is 2 more sizes than MinHeap, pop the top of the MaxHeap heap and push it into MinHeap. if MinHeap is larger than MaxHeap, pop the top of the MinHeap heap and push it into MaxHeap. always keep `|MaxHeap| = |MinHeap| + 1$ or $|MaxHeap| = |MinHeap|$ The

C. Remove the oldest element (window slide):

Retrieve the old_price to be removed from the window queue header.
Reduce the count of old_price in counts.
Delayed deletion: Instead of deleting old_price from the heap immediately, we achieve a delayed and efficient deletion by checking if the top element of the heap counts 0 in counts when getMedian() or balancing the heap, and if it does, popping it until the top element of the heap counts more than 0.

D. getMedian() operation:

Perform cleanup: keep checking the top-of-heap elements of MaxHeap and MinHeap, and if the element has a frequency of 0 in the counts, pop it.
Returns results: due to $K$ is odd, the balanced MaxHeap must be one element larger than MinHeap. Therefore, the median is the top element of the cleaned MaxHeap.

complexity analysis

Time Complexity:
- add(price): both insert and balance are $O(\log K)$ The
- getMedian(): top-of-heap queries are $O(1)$ , the complexity of the worst-case cleanup operation is also evenly distributed $O(\log K)$ The
Space Complexity: $O(K)$ for storing the sliding window in the $K$ elements (in heaps, queues, and hash tables).

this kind of

O(\log K)

The real-time updating and querying efficiency is perfectly acceptable in high-frequency trading scenarios, so it is the optimal solution.

Behavior Questions

Many candidates mistakenly view the Behavioral Interview as "small talk" or a background briefing, but we emphasize at the outset: "Cultural Fit is the easiest elimination point for top companies, so make sure you're there! Make sure you give it your best shot."

Instead of preparing Optiver for his highly competitive and excellence-driven culture, we systematically constructed six high-impact STAR framework stories. These stories accurately cover core competency topics such as dealing with technical challenges, teamwork under pressure, and learning new things at a rapid pace.

He finally selected and rehearsed three sets of stories: one set focuses on how to independently design and drive the data consistency validation program in a complex database migration project, reflecting his extreme attention to detail and responsibility; the other set focuses on the rapid diagnosis and response ability he demonstrated when a temporary unexpected interface anomaly occurred. The interviewer's feedback after listening was very penetrating: "What you told is not a pre-rehearsed script, but your real and deep experience and reflection." This kind of narration, which perfectly combines personal actions with the company's demand for high standards and result orientation, is undoubtedly a huge plus.

Interview Prep

First, the technology must never stop at simple textbook algorithms, but delve deeper into the core challenges in the design of high-frequency, low-latency (HFT) trading systems. In particular, it is important to understand how to design systems with millisecond or even microsecond responses in a trading environment where performance is paramount. For example, it is important to master how to optimize memory access patterns, understand operating system scheduling principles (e.g., interrupts, context switches), and how to ensure data consistency and low latency in concurrent environments to avoid performance bottlenecks caused by lock contention.

Second, the Behavioral and Cultural Fit Interview is critical. Be sure to use the STAR framework to organize your responses, not only describing events, but also reflecting and summarizing in depth "what you learned in an environment of excellence and extreme competition". "Optiver values logical reasoning, problem solving, and the ability to communicate clearly under pressure, and your answers must demonstrate extreme attention to detail and a results-oriented mindset.

Finally, targeted preparation is very efficient. It is recommended that you review each interview, taking notes on the questions, test points, and answer structure, and it is a good idea to obtain a copy of the question bank for Optiver's focus, and to concentrate your efforts on difficult algorithmic questions (especially those involving probability, combinatorial optimization, and deep applications of data structures) and system design questions, especially on understanding design concepts like real-time market data processing, the summarization engine's core principles of such design concepts that are tightly integrated with financial transactions.