TikTok·Machine Learning Engineer·Technical Phone Screen·Senior

SeniorPrefer not to say

Jun 2026Remote

Summary

TikTok ML engineer round with a dynamic programming twist I didn't fully see coming. The problem looked like a standard word break at first glance but the second-smallest cost angle made it genuinely harder.

Questions Asked (1)

Given a string and a dictionary of words, a valid segmentation splits the string into dictionary words whose concatenation equals the original. The cost is the number of words used. Return the second-smallest cost across all valid segmentations, or -1 if fewer than two valid segmentations exist. Also explain your algorithm's time and space complexity and how you'd handle large inputs.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

My first instinct was to track just the minimum cost per index, which is basically the standard word break DP.

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AI HintsAI Generated

Suggested Approach

First, clarify the problem and edge cases, then propose a dynamic programming solution that tracks the two smallest costs for each prefix. Explain how to compute the second-smallest cost efficiently and analyze time/space complexity, including optimizations for large inputs.

Pro tip: Mention that you can use a trie to optimize dictionary lookups and that tracking the two smallest costs avoids storing all segmentations, which is crucial for large inputs.

1. Clarify the problem and edge cases

Restate the problem to ensure understanding, and discuss edge cases such as empty string, no valid segmentation, and exactly one valid segmentation.

2. Design a dynamic programming approach

Define dp[i] as the two smallest costs to segment the prefix of length i. For each i, iterate over all j < i where the substring s[j:i] is in the dictionary, and update dp[i] using dp[j].

3. Optimize dictionary lookups

Use a trie or a hash set for O(1) or O(L) substring lookups, and consider precomputing all valid substrings to avoid repeated checks.

4. Analyze time and space complexity

Time complexity is O(n^2 * L) with naive substring checks, but can be O(n^2) with a trie. Space complexity is O(n) for dp and O(total characters) for the trie.

5. Discuss handling large inputs

For large inputs, use a trie to reduce lookup time, consider memory limits, and possibly use a sliding window or BFS with pruning to avoid unnecessary computations.

Key Points to Mention

Dynamic programming with state tracking two smallest costs
Using a trie or hash set for efficient dictionary lookups
Time complexity: O(n^2 * L) naive, O(n^2) with trie; space complexity: O(n + total dictionary characters)
Handling large inputs: memory optimization, pruning, and avoiding storing all segmentations
Edge cases: empty string, no valid segmentation, exactly one valid segmentation
Returning -1 when fewer than two valid segmentations exist

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