Capital One·Software Engineer·Online Assessment (OA)·Intermediate

IntermediatePrefer not to say

Apr 2026

Summary

Capital One SWE interview with a data structure problem that should've been straightforward but something went sideways with the judge. Left feeling like I had the right idea but couldn't get it to pass.

Questions Asked (1)

You're given a sequence of two operation types: one that places a wall at a specific index, and another that queries whether any wall exists within a given range. For each range query, output 1 if a wall exists in that range and 0 otherwise, then return all results as a list.

Algorithms & Data Structures

Author's notes

My logic felt right.

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AI HintsAI Generated

Suggested Approach

Use a balanced binary search tree (e.g., TreeSet in Java) to store wall indices, allowing O(log n) insertions and range queries via floor/ceiling operations. For each query, check if the smallest wall index >= L is <= R; if so, output 1, else 0. This efficiently handles dynamic updates and range existence checks.

Pro tip: Mention that a Fenwick tree with binary search can also solve this in O(log n) per operation, but a TreeSet is simpler and less error-prone in an interview. Always clarify the expected number of operations and whether indices are bounded to choose the optimal data structure.

1. Clarify requirements and constraints

Ask about the number of operations, index range, and whether walls can be placed at the same index multiple times. This determines if a simple set or a more complex structure is needed.

2. Choose the right data structure

Select a balanced BST (like TreeSet) to maintain sorted wall indices, enabling O(log n) insertions and efficient range queries. Alternatively, consider a Fenwick tree if indices are bounded and updates are frequent.

3. Implement wall placement

For each wall placement operation, insert the index into the data structure. If using a set, duplicates are automatically ignored.

4. Implement range query

For a query [L, R], find the smallest wall index >= L (using ceiling). If it exists and is <= R, output 1; otherwise, output 0.

5. Return results in order

Collect the outputs for each query in a list and return it after processing all operations.

Key Points to Mention

Time complexity: O(log n) per operation with balanced BST, leading to O(m log n) overall.
Space complexity: O(n) for storing wall indices.
Use of ceiling/floor operations for efficient range existence check.
Handling duplicate wall placements gracefully (e.g., set ignores duplicates).
Alternative approaches: Fenwick tree with binary search, segment tree, or bitset if index range is small.
Edge cases: empty range, L > R, no walls placed yet, query range outside placed indices.

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