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

IntermediatePrefer not to say
Jun 2026

Summary

Took an Amazon SDE online assessment and got wrecked by a string DP problem. Scored 6 out of 15 test cases, which is not great, and even after talking it through with others I'm still not sure what the intended solution actually is.

Questions Asked (1)

Q1

Given two strings representing branches of commits, find the minimum number of inversions (conflicts) in any valid interleaving of the two strings, where a conflict is a character with lower alphabetical priority appearing before one with higher priority.

Algorithms & Data Structures
Author's notes

Completely blanked during the actual OA.

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AI HintsAI Generated

Suggested Approach

First, clarify the problem: we need to interleave two strings while minimizing the number of inversions (pairs where a lower-priority character appears before a higher-priority one). Then, propose a dynamic programming solution where the state is the number of characters taken from each string, and the value is the minimum inversions so far. Finally, discuss how to compute the additional inversions when appending a character from one string, considering the characters already taken from the other string.

Pro tip: Mention that this problem is equivalent to finding a minimum-cost path in a grid, and that the cost of adding a character can be precomputed using prefix sums to achieve O(n*m) time. Also, note that if the strings are large, we can optimize space to O(min(n,m)) by using a 1D DP array.

1. Clarify the problem and define inversion

Confirm that an inversion is a pair (i, j) with i < j in the interleaved string where the character at i has lower alphabetical priority than the character at j. Also, clarify that 'lower alphabetical priority' means a character that comes earlier in the alphabet (e.g., 'a' has lower priority than 'b').

2. Define DP state and transition

Let dp[i][j] be the minimum inversions in an interleaving of the first i characters of string A and the first j characters of string B. When appending A[i] to the interleaving, the additional inversions are the number of characters in B[0..j-1] that have higher priority than A[i] (i.e., are alphabetically greater). Similarly for appending B[j].

3. Precompute additional inversion costs

Precompute for each character in A and each prefix of B, the count of characters in that prefix that are greater than the character. Similarly for B and prefixes of A. This can be done with prefix sums over the alphabet (26 letters) for O(1) lookup.

4. Implement DP and handle base cases

Initialize dp[0][0] = 0. For i from 0 to n, for j from 0 to m, update dp[i+1][j] and dp[i][j+1] using the precomputed costs. The answer is dp[n][m]. Discuss time and space complexity: O(n*m) time, O(n*m) space, with possible optimization to O(min(n,m)) space.

5. Test with examples and edge cases

Walk through a small example (e.g., A='ab', B='ba') to verify the DP. Consider edge cases: empty strings, identical strings, strings with all characters in increasing or decreasing order.

Key Points to Mention

  • Dynamic programming with state (i, j) representing the number of characters taken from each string.
  • The cost of appending a character is the number of characters already taken from the other string that have higher alphabetical priority.
  • Precomputation of costs using prefix sums over the alphabet to achieve O(1) per transition.
  • Time complexity O(n*m) and space complexity O(n*m), with possible optimization to O(min(n,m)) space.
  • The problem is equivalent to finding a minimum-cost path in a grid from (0,0) to (n,m).
  • Edge cases: empty strings, strings with no inversions, and strings with maximum inversions.

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