Databricks·Software Engineer·Technical Phone Screen·Senior

SeniorPrefer not to say

Jun 2026Remote

Summary

Databricks coding round, one problem the whole time. It was a string compression thing with DP and honestly more involved than it sounds at first glance.

Questions Asked (1)

Implement a string compressor using run-length encoding, but use dynamic programming over the string's prefixes to find the optimal segmentation that minimizes the total encoded length. Account for encoding overhead (e.g. a count of 1 might be written as just the character, not 'c1'). Return either the minimum encoded length or the actual compressed string.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I knew RLE from the classic interview version and jumped straight into a greedy pass.

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AI HintsAI Generated

Suggested Approach

Define a DP where dp[i] is the minimum encoded length for the prefix of length i, and for each i consider all possible last segments ending at i, computing the encoded length of each segment and taking the minimum. To reconstruct the string, store the chosen segment start and encoded representation for each i. Handle the count-of-1 special case by encoding a single character as just the character.

Pro tip: Clarify upfront whether the output should be the minimum length or the actual compressed string, and mention that the DP can be extended to reconstruct the string with parent pointers. Also, discuss the trade-off between O(n^2) DP and possible optimizations like limiting segment length or using suffix automata for large inputs.

1. Define DP state and base case

Let dp[i] be the minimum encoded length for the prefix of length i. Base case dp[0] = 0. For each i from 1 to n, compute dp[i] by considering all j < i.

2. Compute encoded length of a segment

For a segment s[j:i], compute its run-length encoding length: if the segment is a single character, length is 1; otherwise, sum over runs of (1 + number of digits in run length).

3. Transition and optimize

For each i, iterate j from 0 to i-1, compute segment cost, and update dp[i] = min(dp[i], dp[j] + cost). This yields O(n^2) time and O(n) space.

4. Reconstruct the compressed string (if needed)

Store the chosen j for each i (e.g., in a parent array). After computing dp[n], backtrack from n to 0 to build the compressed string by concatenating the encoded segments.

5. Analyze complexity and discuss trade-offs

Time complexity is O(n^2) due to nested loops; space is O(n). Discuss potential optimizations like limiting segment length or using more advanced data structures for large n.

Key Points to Mention

Dynamic programming over prefixes with optimal substructure.
Run-length encoding and the special case for count of 1 (omit the count).
Computing the encoded length of a segment efficiently, including digit counting.
Reconstruction of the compressed string using parent pointers.
Time and space complexity analysis (O(n^2) time, O(n) space).
Trade-offs and potential optimizations for large inputs.

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