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Trexquant·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
May 2026

Summary

C++ fundamentals screening at Trexquant for a software engineer role. After a coding question, the interviewer drilled into language mechanics, specifically pointer const-correctness and container choice. Nothing outrageous, but the follow-ups got detailed fast.

Questions Asked (4)

Q1

What is the difference between `const int*` and `int* const` in C++? What exactly is immutable in each case, and how do you read declarations like `const int* const` in general?

Technical Trade-offsAlgorithms & Data Structures
Author's notes

I knew the surface answer but fumbled when asked to be precise about which assignment fails to compile.

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AI HintsAI Generated

Suggested Approach

Start by explaining the general rule for reading declarations: start at the variable name and work outward, applying const to the nearest type to its left (or right if nothing to the left). Then apply this to `const int*` (pointer to const int) and `int* const` (const pointer to int), clarifying what is immutable in each case. Finally, extend to `const int* const` (const pointer to const int) and mention the alternative syntax `int const *` for consistency.

Pro tip: Use the spiral rule or the 'const applies to the thing on its left, unless there's nothing on its left, then it applies to the thing on its right' mnemonic. Also, mention that `const int*` and `int const*` are equivalent, and that `int* const` requires initialization.

1. Explain the general rule for reading declarations

Describe the clockwise/spiral rule or the 'const applies to the left' rule. Start at the variable name and move outward, applying const to the nearest type to its left (or right if nothing to the left).

2. Analyze `const int*`

Explain that this is a pointer to a const int. The int value cannot be modified through the pointer, but the pointer itself can be reassigned to point to another const int.

3. Analyze `int* const`

Explain that this is a const pointer to an int. The pointer cannot be reassigned after initialization, but the int value it points to can be modified.

4. Analyze `const int* const`

Explain that this is a const pointer to a const int. Neither the pointer nor the pointed-to value can be modified. Mention that the pointer must be initialized.

5. Summarize and provide examples

Summarize the differences and give code examples to illustrate. Mention that `const int*` and `int const*` are equivalent, and that `int* const` requires initialization.

Key Points to Mention

  • `const int*` is a pointer to a const int; the int cannot be modified through the pointer, but the pointer can be reassigned.
  • `int* const` is a const pointer to an int; the pointer cannot be reassigned, but the int can be modified.
  • `const int* const` is a const pointer to a const int; neither the pointer nor the int can be modified.
  • The general rule: const applies to the type immediately to its left, unless there is nothing to its left, then it applies to the type to its right.
  • `const int*` and `int const*` are equivalent; `int* const` is different.
  • `int* const` must be initialized at declaration, while `const int*` does not require initialization.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.

Q2

Explain the difference between `std::map` and `std::unordered_map`. Cover the underlying data structures, operation complexities, ordering guarantees, key type requirements, and when you'd pick one over the other.

Technical Trade-offsSystem Design
Author's notes

This felt more comfortable.

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AI HintsAI Generated

Suggested Approach

Structure your answer by comparing the two containers across the five dimensions: underlying data structure, operation complexities, ordering, key requirements, and use cases. Emphasize the trade-offs and give concrete examples of when each is preferable, especially in performance-critical contexts like trading systems.

Pro tip: Mention that std::unordered_map has average O(1) but worst-case O(n) due to hash collisions, and that in latency-sensitive systems like trading, the predictability of std::map's O(log n) can be preferable to avoid tail latency spikes.

1. Underlying Data Structures

Explain that std::map is typically implemented as a balanced binary search tree (e.g., red-black tree), while std::unordered_map uses a hash table.

2. Operation Complexities

Compare average and worst-case time complexities for insertion, deletion, and lookup: std::map is O(log n) for all, std::unordered_map is average O(1) but worst-case O(n).

3. Ordering Guarantees

State that std::map maintains elements in sorted order by key, while std::unordered_map has no defined order.

4. Key Type Requirements

Note that std::map requires keys to be comparable with operator< (or a custom comparator), while std::unordered_map requires keys to be hashable and equality-comparable.

5. Choosing Between Them

Discuss scenarios: use std::map when order matters or when you need predictable performance; use std::unordered_map when average-case speed is critical and order is irrelevant.

Key Points to Mention

  • std::map is a balanced BST (red-black tree), std::unordered_map is a hash table.
  • std::map operations are O(log n); std::unordered_map average O(1), worst-case O(n).
  • std::map keeps keys sorted; std::unordered_map has no ordering.
  • std::map requires operator< or comparator; std::unordered_map requires std::hash and operator==.
  • Use std::map for ordered traversal, range queries, or when worst-case guarantees matter.
  • Use std::unordered_map for fast average-case lookups when order is not needed.
  • Consider memory overhead and cache locality: std::map has higher per-element overhead; std::unordered_map may have better locality but rehashing costs.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.

Q3

If you needed keys in sorted order AND fast average-case lookups, how would you approach that? What are the trade-offs of maintaining both a map and an unordered_map versus sorting on demand?

Technical Trade-offsSystem Design
Author's notes

Surprised me a bit as a follow-up.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the requirements: what operations are needed (insert, lookup, iteration in sorted order), expected data size, and performance constraints. Then propose a dual-structure approach (e.g., hash map + balanced BST or sorted array) and discuss the trade-offs versus sorting on demand, including time/space complexity and update patterns.

Pro tip: Emphasize that the best choice depends on the read/write ratio and whether sorted order is needed frequently or just occasionally; mention that in practice, a single ordered structure like a balanced BST or skip list can provide both O(log n) lookups and sorted iteration, avoiding the overhead of maintaining two structures.

1. Clarify Requirements

Ask about the expected operations (insert, delete, lookup, sorted iteration), data size, and performance requirements (e.g., latency, throughput).

2. Evaluate Dual-Structure Approach

Propose maintaining both a hash map for O(1) average lookups and an ordered structure (e.g., balanced BST, skip list, or sorted array) for sorted iteration. Discuss synchronization and memory overhead.

3. Evaluate Sort-on-Demand Approach

Consider using only a hash map and sorting keys when needed. Analyze the cost: O(n log n) per sort, which may be acceptable if sorted access is infrequent.

4. Compare Trade-offs

Compare time complexity (lookup, insert, sorted iteration), space usage, and code complexity. Consider update frequency and whether sorted order is needed often.

5. Recommend and Justify

Choose an approach based on the clarified requirements and justify it with concrete reasoning, mentioning alternatives like using a single ordered map (e.g., std::map) if O(log n) lookups are acceptable.

Key Points to Mention

  • Time complexity: O(1) average lookup for hash map, O(log n) for balanced BST, O(n log n) for sorting on demand.
  • Space overhead: dual structures require more memory and careful synchronization.
  • Update patterns: frequent inserts/deletes make maintaining two structures costly; sorting on demand may be better if updates are batched.
  • Alternative: use a single ordered structure (e.g., std::map, skip list) that provides both O(log n) lookups and sorted iteration.
  • Caching sorted order: if sorted access is repeated, cache the sorted keys and invalidate on updates.
  • Real-world considerations: concurrency, persistence, and library support (e.g., Python's dict + sorted() vs. SortedDict).

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.

Q4

How could an adversary degrade an `unordered_map` to O(n) per operation, and how do you defend against it?

Technical Trade-offsAlgorithms & Data Structures
Author's notes

Hash collision attacks, basically crafting keys that all land in the same bucket.

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AI HintsAI Generated

Suggested Approach

Explain that hash table degradation occurs when many keys collide into the same bucket, turning operations into linear scans. Describe how an adversary can force collisions by exploiting predictable hash functions or by crafting keys that hash to the same value, and then outline defenses such as randomized hashing, balanced tree fallback, and load factor management.

Pro tip: Mention that real-world systems like Java's HashMap and C++'s std::unordered_map have specific mitigations (treeification and prime bucket counts) and that understanding these shows depth beyond textbook knowledge.

1. Explain the degradation mechanism

Describe how hash collisions cause multiple keys to map to the same bucket, and if all keys collide, operations become O(n) because the bucket's linked list must be traversed.

2. Describe adversarial attack

Explain that an adversary who knows the hash function can generate many keys with the same hash, forcing worst-case behavior and potentially causing denial-of-service.

3. Discuss defenses: randomized hashing

Mention using a random seed per execution (e.g., SipHash) to make hash values unpredictable, preventing precomputed collision attacks.

4. Discuss defenses: balanced tree fallback

Explain that some implementations (e.g., Java 8+ HashMap) convert buckets to balanced trees (e.g., red-black trees) when collisions exceed a threshold, guaranteeing O(log n) worst-case per operation.

5. Mention other mitigations and trade-offs

Bring up load factor tuning, prime bucket counts, and monitoring for collision attacks; note that these defenses add overhead and complexity.

Key Points to Mention

  • Hash collisions and worst-case O(n) behavior in hash tables
  • Adversarial key generation to force collisions (e.g., hash flooding)
  • Randomized hashing (e.g., SipHash) to prevent predictability
  • Treeification of buckets (e.g., Java's HashMap) for O(log n) worst-case
  • Load factor and resizing strategies to maintain performance
  • Trade-offs: overhead of randomization and tree structures vs. security

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.