Snowflake·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

May 2026Remote

Summary

Snowflake SWE interview with a coding round that was basically the Dutch National Flag problem dressed up in a different costume. The API-only constraint made it trickier than it looks on paper.

Questions Asked (1)

You're given a RecordCollection object with size(), getColor(i), and swap(i, j) as the only available operations. Each record has a color: 0 (Red), 1 (Green), or 2 (Blue). Sort the collection in place so all reds come first, then greens, then blues. Must run in O(n) time, O(1) space, single pass, no auxiliary arrays.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The moment I saw the three-color constraint and single-pass requirement I knew it was the classic three-pointer partition problem.

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AI HintsAI Generated

Suggested Approach

Recognize this as the Dutch National Flag problem and propose a three-pointer partitioning approach. Explain how to maintain three regions (red, green, blue) in a single pass using swaps, ensuring O(n) time and O(1) space. Walk through the algorithm with a small example to demonstrate correctness.

Pro tip: Emphasize that the algorithm must handle edge cases like all elements the same color or empty collection, and that the swap operation is the only way to modify the collection, so you must carefully manage pointer updates to avoid infinite loops.

1. Clarify constraints and confirm understanding

Restate the problem: sort in-place with only size(), getColor(i), and swap(i, j), in O(n) time, O(1) space, single pass. Confirm that colors are 0,1,2 and that the collection is mutable via swap.

2. Identify the algorithm pattern

Recognize this as the Dutch National Flag problem, which uses three pointers to partition the array into three regions in one pass.

3. Define pointers and invariants

Use low, mid, and high pointers. Invariant: [0, low-1] are reds, [low, mid-1] are greens, [mid, high] are unknown, [high+1, n-1] are blues. Initialize low=0, mid=0, high=n-1.

4. Walk through the algorithm

While mid <= high: if getColor(mid) == 0, swap(low, mid), low++, mid++; if == 1, mid++; if == 2, swap(mid, high), high-- (do not increment mid). Explain why this maintains invariants and terminates.

5. Analyze complexity and edge cases

Each element is examined at most once, so O(n) time. Only constant extra space for pointers. Discuss edge cases: empty collection, all same color, and already sorted.

Key Points to Mention

Dutch National Flag algorithm (three-way partitioning)
Three pointers: low, mid, high with clear invariants
Single pass through the collection
In-place swaps only, no auxiliary arrays
Time complexity O(n), space complexity O(1)
Handling of edge cases (empty, all same color, already sorted)

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