← NVIDIA Interview Insights

NVIDIA·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
May 2026

Summary

NVIDIA coding interview with a classic grouping problem. Nothing too wild but it's the kind of question where your first instinct can lead you down a slow path if you're not careful.

Questions Asked (1)

Q1

Given an array of strings, group all anagrams together. Two strings are anagrams if they share the same characters at the same frequencies. Order of the groups doesn't matter.

Algorithms & Data Structures
Author's notes

My first move was sorting each string and using that as a hash map key, which works fine.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Clarify assumptions and edge cases, then propose a hash map approach where the key is a canonical representation of each string's character frequencies (e.g., sorted string or count tuple). Walk through the algorithm, analyze time/space complexity, and discuss potential optimizations or trade-offs.

Pro tip: Mention that sorting each string to form the key takes O(k log k) per string, but using a character count array as the key can achieve O(k) per string, which is more efficient for large alphabets or long strings. Also, note that the order of groups doesn't matter, so you can return the hash map values directly.

1. Clarify and Define

Confirm input constraints (e.g., string length, character set, empty strings) and output format. Define what makes two strings anagrams.

2. Choose Canonical Key

Decide on a canonical representation for anagrams: sorted string or frequency count array/tuple. Discuss trade-offs in time and space.

3. Hash Map Grouping

Iterate through the array, compute the key for each string, and append the string to the corresponding list in a hash map.

4. Return Groups

Return the values of the hash map as the grouped anagrams. Order doesn't matter.

5. Analyze Complexity

State time complexity: O(n * k) if using count array, O(n * k log k) if sorting. Space complexity: O(n * k) for storing all strings.

Key Points to Mention

  • Hash map with canonical key (sorted string or frequency count)
  • Time complexity analysis: O(n * k) vs O(n * k log k)
  • Space complexity: O(n * k) for output storage
  • Handling edge cases: empty strings, single-character strings, duplicates
  • Alternative approaches: sorting the array first? Not efficient.
  • Optimization: use character count array as key for O(k) per string

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