Amazon·Software Engineer·Online Assessment (OA)·Intermediate

Intermediate

May 2026

Summary

Amazon SWE online assessment with a string partitioning problem that looks deceptively approachable until you realize the edge cases pile up fast. Nothing behavioral, just pure algorithmic grind.

Questions Asked (1)

Given a string of uppercase letters, for each prefix of length i, determine whether the entire string can be split into consecutive blocks of length i where every block has the exact same 26-letter frequency vector as that prefix. Output 0 if valid, 1 otherwise, for each prefix length.

Algorithms & Data Structures

Author's notes

The frequency signature idea clicked pretty quickly but I kept second-guessing myself on the edge cases.

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AI HintsAI Generated

Suggested Approach

Use prefix frequency vectors and check divisibility: for each prefix length i, if n % i != 0, output 1; otherwise, compare the frequency vector of the first i characters with the frequency vector of each subsequent block of length i. To optimize, precompute prefix frequency arrays or use rolling hashes of frequency vectors to enable O(1) block comparisons.

Pro tip: Mention that you can precompute a rolling hash of the frequency vector for each prefix to compare blocks in O(1) time, reducing the overall complexity to O(n log n) or O(n sqrt n) depending on implementation. Also, note that you only need to check prefix lengths that divide n.

1. Understand the problem and constraints

Clarify that for each prefix length i (1 to n), you must determine if the string can be partitioned into blocks of length i, each having the same frequency vector as the prefix of length i. Note that only lengths dividing n can be valid.

2. Precompute frequency vectors

Compute the frequency vector of the entire string and of each prefix. This can be done by maintaining a running count of each letter as you iterate through the string.

3. Check divisibility and compare blocks

For each i from 1 to n, if n % i != 0, mark as invalid. Otherwise, compare the frequency vector of the first i characters with the frequency vector of each subsequent block of length i. If all match, mark valid.

4. Optimize with hashing or early termination

To avoid O(n^2) comparisons, use a rolling hash of frequency vectors or precompute prefix frequency arrays to compare blocks in O(1). Alternatively, break early if a mismatch is found.

5. Output results and analyze complexity

Return an array of 0s and 1s for each prefix length. Discuss time and space complexity, and possible optimizations for large n.

Key Points to Mention

Frequency vector representation (e.g., array of 26 integers)
Divisibility condition: only consider i where n % i == 0
Efficient comparison of frequency vectors using hashing or prefix sums
Time complexity analysis: naive O(n^2) vs optimized O(n log n) or O(n sqrt n)
Edge cases: i = n (always valid if string non-empty), i = 1 (valid only if all characters same)
Space-time tradeoff: precomputing prefix frequency arrays uses O(26n) space but allows O(1) block comparisons

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