Oracle·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Oracle SWE interview with a sliding window coding problem. Nothing too surprising but it required knowing your character frequency tricks cold.

Questions Asked (1)

Given two strings s and p, find all starting indices in s where a substring is an anagram of p.

Algorithms & Data Structures

Author's notes

Classic sliding window.

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AI HintsAI Generated

Suggested Approach

Use a sliding window of length equal to p's length, maintaining frequency counts of characters in the window and comparing them to p's frequency counts. Optimize by updating counts incrementally as the window slides, rather than recomputing from scratch. Return all starting indices where the window's character counts match p's.

Pro tip: Mention that you can avoid comparing full frequency arrays each time by tracking the number of matches between the two count arrays, updating it in O(1) per slide. This shows you understand constant-factor optimizations and can discuss trade-offs between simplicity and performance.

1. Clarify constraints and edge cases

Ask about input size, character set (e.g., lowercase English letters), and whether p can be longer than s. Discuss handling empty strings or no matches.

2. Choose the sliding window approach

Explain that a fixed-size window of length p.length() slides over s, and we compare character frequencies. This avoids checking all substrings naively.

3. Implement frequency counting and comparison

Use arrays or hash maps to count characters in p and in the current window. Compare counts efficiently, e.g., by tracking matches or using a counter.

4. Slide the window and update counts

Move the window one step: add the new character on the right, remove the old character on the left, and update the match count accordingly.

5. Collect and return results

Whenever the window's counts match p's, record the starting index. After processing, return the list of indices.

Key Points to Mention

Time complexity: O(n) where n is length of s, with constant factor depending on alphabet size.
Space complexity: O(1) if using fixed-size arrays for lowercase letters, or O(k) for hash maps where k is distinct characters.
Edge cases: p longer than s, empty strings, no anagrams present.
Optimization: maintain a 'matches' counter to avoid O(alphabet) comparison per slide.
Alternative approaches: sorting each substring (O(n * m log m)) is less efficient; mention but explain why sliding window is better.
Handling Unicode or larger character sets: use hash maps instead of arrays.

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