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Amazon·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
Jun 2026

Summary

Amazon SWE coding round, one problem the whole session. Classic dynamic programming territory but with a twist on the output format that tripped me up a bit.

Questions Asked (1)

Q1

Given a string and a dictionary of words, determine if the string can be segmented into a sequence of dictionary words (reuse allowed). If yes, return one valid decomposition as an ordered list; if not, return an empty list. Discuss time and space complexity for large inputs.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

I knew word break immediately, done it before.

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AI HintsAI Generated

Suggested Approach

Use dynamic programming to determine if the string can be segmented, then backtrack to reconstruct one valid decomposition. Start by defining a DP array where dp[i] indicates if the prefix of length i is segmentable, and populate it by checking all possible last words. For reconstruction, store the starting index of the last word for each reachable position, then follow the links backward to build the list.

Pro tip: After presenting the DP solution, mention that for very large inputs, you can optimize by limiting the maximum word length considered or using a trie to reduce unnecessary checks, showing awareness of practical scalability.

1. Clarify and Define

Restate the problem to ensure understanding, including constraints like reuse allowed and returning any valid decomposition. Ask about input size, character set, and dictionary size to guide complexity analysis.

2. Design DP for Feasibility

Define dp[i] as whether the prefix of length i can be segmented. Initialize dp[0] = true, and for each i from 1 to n, check all j < i where dp[j] is true and substring(j, i) is in the dictionary.

3. Reconstruct One Solution

During DP, store the starting index of the last word for each reachable i (e.g., in a parent array). After filling DP, if dp[n] is true, backtrack from n to 0 using the parent array to build the list of words.

4. Analyze Complexity

Time complexity is O(n^2 * L) where n is string length and L is average word length for substring checks, or O(n * m) if using a trie with m as max word length. Space complexity is O(n) for DP and parent arrays, plus O(n) for the result.

5. Discuss Optimizations and Trade-offs

Mention optimizations like using a trie for faster lookups, limiting j to max word length, or using BFS/DFS with memoization. Discuss trade-offs between time and space, and when to choose one approach over another.

Key Points to Mention

  • Dynamic programming formulation with dp[i] indicating segmentability of prefix i.
  • Reconstruction using a parent array to store the start index of the last word.
  • Time complexity: O(n^2 * L) naive, O(n * m) with trie optimization, where n is string length, L is average word length, m is max word length.
  • Space complexity: O(n) for DP and parent arrays, plus O(n) for output.
  • Handling edge cases: empty string, empty dictionary, no valid segmentation.
  • Optimization techniques: trie for dictionary, limiting substring checks to max word length, BFS/DFS with memoization.

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