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NVIDIA·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jul 2026

Summary

Nvidia SWE interview with a classic in-place array sorting problem. Nothing too wild but the constraints matter more than people expect.

Questions Asked (1)

Q1

Given an array of n elements each representing one of three colors (red, white, or blue), sort the array in-place so all same-color elements are grouped together in the order red, white, then blue.

Algorithms & Data Structures
Author's notes

The naive approach works but they clearly wanted something better.

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AI HintsAI Generated

Suggested Approach

Use the Dutch National Flag algorithm (three-pointer approach) to partition the array into three sections: red, white, and blue. Maintain three pointers: low for the boundary of red, mid for the current element, and high for the boundary of blue. Iterate through the array, swapping elements to place them in the correct section, achieving O(n) time and O(1) space.

Pro tip: Emphasize that this is a one-pass solution with constant space, which is optimal for large datasets. Mention that the algorithm is stable in terms of color grouping but not necessarily preserving original order within colors, which is acceptable here.

1. Clarify the problem and constraints

Confirm that the array contains only three distinct values (e.g., 0, 1, 2) and that sorting must be in-place. Ask if stability within colors is required (usually not).

2. Choose the algorithm

Select the Dutch National Flag algorithm (three-pointer approach) as it optimally solves the problem in O(n) time and O(1) space.

3. Initialize pointers

Set low = 0, mid = 0, and high = n-1. These pointers divide the array into four regions: red (0 to low-1), white (low to mid-1), unknown (mid to high), and blue (high+1 to n-1).

4. Iterate and swap

While mid <= high, examine the element at mid. If it's red (0), swap with low and increment both low and mid. If white (1), just increment mid. If blue (2), swap with high and decrement high (do not increment mid).

5. Verify and discuss complexity

After the loop, the array is sorted. Explain that each element is examined at most once, giving O(n) time, and only a few pointers are used, giving O(1) space.

Key Points to Mention

  • Dutch National Flag algorithm (three-pointer technique)
  • Time complexity: O(n) single pass
  • Space complexity: O(1) in-place
  • Handling of edge cases (empty array, all same color)
  • Comparison with counting sort (two-pass) and why one-pass is better
  • Stability not required, so swaps are acceptable

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