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Google·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
Apr 2026

Summary

Google SWE coding round, got a dynamic programming string problem. Pretty standard stuff but the edge cases are where you can lose points if you're not careful.

Questions Asked (1)

Q1

Given a string and a dictionary of words, determine whether the string can be fully segmented into a sequence of dictionary words.

Algorithms & Data Structures
Author's notes

Classic dp problem once you see it, but my first instinct was to go recursive without memoization and I had to backtrack.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem constraints (e.g., can words be reused, case sensitivity, empty string). Then propose a dynamic programming solution where dp[i] indicates if the prefix of length i can be segmented, and for each i, check all dictionary words that end at i. Optimize with a trie or BFS to avoid redundant checks.

Pro tip: After presenting the DP solution, mention that you can optimize space to O(n) and time to O(n * maxWordLength) by using a trie or by iterating over word lengths present in the dictionary. Also, discuss trade-offs between DP and BFS/DFS with memoization.

1. Clarify constraints and edge cases

Ask about input size, character set, whether words can be reused, and if the dictionary is static. Handle edge cases like empty string, empty dictionary, and very long strings.

2. Define the DP state and recurrence

Define dp[i] as whether the substring s[0..i-1] can be segmented. Initialize dp[0] = true. For each i from 1 to n, dp[i] = true if there exists j < i such that dp[j] is true and s[j..i-1] is in the dictionary.

3. Optimize the transition

Instead of checking all j, iterate over dictionary words and check if they match a suffix ending at i. Alternatively, use a trie to efficiently find all valid words ending at i, or use BFS with memoization to avoid redundant subproblems.

4. Analyze complexity and discuss trade-offs

Naive DP is O(n^2 * L) where L is average word length. With a trie or by limiting to max word length, it becomes O(n * maxWordLength). Space is O(n). Compare with BFS/DFS approaches.

5. Test with examples and edge cases

Walk through examples like 'leetcode' with ['leet','code'] and 'applepenapple' with ['apple','pen']. Also test cases where segmentation is impossible, e.g., 'catsandog' with ['cats','dog','sand','and','cat'].

Key Points to Mention

  • Dynamic programming with boolean array dp where dp[i] indicates if prefix of length i is segmentable.
  • Optimization using a trie to efficiently check if a substring is a dictionary word.
  • Time and space complexity analysis, including worst-case and optimized versions.
  • Handling of edge cases: empty string, empty dictionary, overlapping words, and repeated words.
  • Alternative approaches: BFS/DFS with memoization, and their trade-offs.
  • Clarifying questions about input constraints and dictionary properties.

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