Databricks·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Apr 2026

Summary

Databricks coding round, one question but it had a twist I didn't fully see coming. The base version of the problem I knew, but they pushed further and that's where things got interesting.

Questions Asked (1)

Given two strings s1 and s2, find all starting indices in s2 where the substring of length |s1| is an anagram (permutation) of s1. Return every such index.

Algorithms & Data Structures

Author's notes

I knew the permutation-in-string problem so the setup felt familiar, but returning all matches instead of just true/false tripped me up for a minute.

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AI HintsAI Generated

Suggested Approach

Use a sliding window of length |s1| over s2, maintaining frequency counts of characters in the window and comparing them to the frequency counts of s1. To optimize, use a fixed-size array of 26 integers (for lowercase English letters) and update counts incrementally as the window slides, achieving O(n) time.

Pro tip: Mention that you can avoid comparing full frequency arrays each time by tracking the number of characters whose counts match, reducing the comparison to O(1) per window. Also, clarify assumptions about character set (e.g., lowercase English letters) and handle edge cases like empty strings or s1 longer than s2.

1. Clarify and Validate Input

Confirm the character set (e.g., lowercase English letters) and edge cases: if s1 is longer than s2, return empty; if either is empty, decide on behavior. This shows attention to detail.

2. Choose Data Structures

Use a fixed-size frequency array (size 26) for s1 and for the sliding window. Alternatively, use a hash map for general character sets, but arrays are more efficient for known small alphabets.

3. Initialize the First Window

Compute frequency counts for the first |s1| characters of s2 and compare with s1's counts. If they match, add index 0 to the result.

4. Slide the Window

For each subsequent index i from |s1| to len(s2)-1, add the new character at i and remove the character at i-|s1| from the window counts. Compare updated counts with s1's counts; if equal, add the start index (i-|s1|+1) to the result.

5. Optimize Comparison

Instead of comparing entire arrays each time, maintain a 'matches' counter that tracks how many characters have the correct frequency. Update it incrementally when adding/removing characters, allowing O(1) comparison per window.

Key Points to Mention

Sliding window technique for O(n) time complexity
Frequency counting using arrays or hash maps
Incremental updates to avoid recomputing counts from scratch
Optimization: tracking number of matching characters to achieve O(1) comparison per window
Edge cases: s1 longer than s2, empty strings, and character set assumptions
Space complexity: O(1) if using fixed-size array for lowercase letters, O(k) for hash map with k distinct characters

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