Databricks·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Databricks coding screen, one question, sliding window. Pretty standard stuff but worth writing up since the problem has a few gotchas if you're not warmed up on frequency maps.

Questions Asked (1)

Given two strings, determine whether any permutation of the first string appears as a substring of the second string.

Algorithms & Data Structures

Author's notes

Classic sliding window with a character frequency array.

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AI HintsAI Generated

Suggested Approach

Clarify the problem and edge cases, then propose a sliding window with character frequency counting to check if any substring of the second string is a permutation of the first. Discuss time and space complexity, and consider follow-up optimizations or alternative approaches.

Pro tip: Mention that you can optimize the comparison of frequency arrays by maintaining a 'matches' count to avoid O(26) checks per window, and discuss how this scales for larger character sets.

1. Clarify the problem

Ask about input constraints: string lengths, character set (e.g., lowercase English letters), and whether case sensitivity matters. Confirm that 'permutation' means same characters with same frequencies, order irrelevant.

2. Outline the approach

Propose using a sliding window of length equal to the first string over the second string. Maintain frequency counts of characters in the window and compare with the frequency count of the first string.

3. Detail the algorithm

Initialize frequency arrays for the first string and the first window. Slide the window one character at a time: add the new character, remove the old character, and check if the frequency arrays match. If they match, return true.

4. Analyze complexity

State that the time complexity is O(n) where n is the length of the second string, assuming constant alphabet size (e.g., 26). Space complexity is O(1) for fixed alphabet, or O(k) for k distinct characters.

5. Discuss edge cases and optimizations

Handle cases where the first string is longer than the second, empty strings, and repeated characters. Mention optimization: maintain a 'matches' count to avoid comparing entire frequency arrays each time.

Key Points to Mention

Sliding window technique with fixed window size
Character frequency counting (e.g., using arrays or hash maps)
Time complexity O(n) and space complexity O(1) for fixed alphabet
Edge cases: empty strings, first string longer than second, repeated characters
Optimization: maintain a 'matches' count to achieve O(1) comparison per window
Alternative approaches: sorting each substring (O(n * m log m)) or using a hash map for variable character sets

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