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

Intermediate
Jun 2026

Summary

Amazon SWE online assessment with a pretty nasty string DP problem. The kind of question where you think you understand it, then realize you've been solving the wrong thing for 20 minutes.

Questions Asked (1)

Q1

You're given a string of '0's, '1's, and '!' characters, where each '!' can be replaced with either '0' or '1'. After replacement, compute count10 (number of subsequence pairs where '1' comes before '0') and count01 (pairs where '0' comes before '1'). Given weights x and y, maximize x * count10 + y * count01 over all possible replacements, modulo 1e9+7.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

The problem looks deceptively clean until you realize the '!' replacements interact with each other and you can't just greedily assign them left to right without thinking about what's already been placed.

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AI HintsAI Generated

Suggested Approach

First, clarify the problem and constraints, then derive a dynamic programming solution that processes the string left-to-right, tracking the number of 0s and 1s seen so far and the contribution of each character to the counts. For '!' characters, decide greedily based on the weights x and y, but ensure the DP state captures enough information to handle dependencies. Finally, implement the DP with modulo arithmetic and analyze time and space complexity.

Pro tip: Mention that the greedy choice for '!' depends on the difference between x and y and the current counts, and that a DP with state (number of 0s, number of 1s) can be optimized to O(n) by observing that only the difference matters. Also, discuss potential pitfalls like integer overflow and modulo handling.

1. Clarify the problem and constraints

Restate the problem in your own words, ask about input size, modulo, and edge cases (e.g., all '!', empty string). Confirm that subsequences are not necessarily contiguous.

2. Derive the contribution of each character

For a fixed string, count10 is the sum over each '1' of the number of '0's after it, and count01 is the sum over each '0' of the number of '1's after it. This can be computed in one pass by maintaining counts of 0s and 1s seen so far.

3. Handle '!' with dynamic programming

Process characters left-to-right. Maintain DP states representing the number of 0s and 1s seen so far, and the accumulated weighted counts. For '!', branch on assigning 0 or 1, updating counts and adding contributions. Use modulo arithmetic.

4. Optimize the DP

Observe that the contribution of a new character depends only on the difference between the number of 0s and 1s seen so far, not both individually. Reduce the state to the difference, which ranges from -n to n, and use a 1D DP array. This yields O(n^2) time, which can be further optimized to O(n) with greedy insights if applicable.

5. Analyze complexity and trade-offs

Discuss time and space complexity of the DP and any greedy alternative. Explain why the greedy choice for '!' might be optimal: if x > y, we prefer 1s early and 0s late; if x < y, the opposite. Compare with DP for correctness.

Key Points to Mention

  • Definition of count10 and count01 as subsequence pairs, not substrings.
  • One-pass computation of counts for a fixed string.
  • Dynamic programming state: difference between number of 0s and 1s, and accumulated weighted sum.
  • Greedy decision for '!' based on weights x and y: if x > y, assign '1' when possible to maximize count10; if x < y, assign '0' to maximize count01.
  • Modulo arithmetic to prevent overflow and meet problem constraints.
  • Time and space complexity: O(n^2) DP, potential O(n) greedy solution, and trade-offs.

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