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Snowflake·Data Scientist·Onsite - Coding / Algorithms·Senior

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Jul 2026

Summary

Snowflake Data Scientist interview that went deep into algorithmic problem solving, way deeper than I expected for a DS role. The whole session was essentially a multi-part coding and system design marathon centered on a single problem with escalating constraints. Left feeling like I'd been wrung out.

Questions Asked (4)

Q1

Implement an O(n^2) solution to find all unique triplets in an array that sum to zero, using sorting and two pointers. Prove your duplicate-skipping logic is correct on adversarial inputs like an array of all zeros.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

The basic two-pointer approach I knew cold.

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AI HintsAI Generated

Suggested Approach

Start by explaining the sort + two-pointer algorithm: sort the array, then for each index i, use two pointers to find pairs that sum to -nums[i]. Emphasize the duplicate-skipping conditions for i and for the pointers, and prove correctness by analyzing edge cases like all zeros.

Pro tip: When discussing duplicate skipping, explicitly state the invariant that after skipping, the next element is strictly greater than the previous, ensuring no duplicate triplets. Also, mention that the algorithm is O(n^2) because sorting is O(n log n) and the two-pointer scan is O(n^2).

1. Sort the array

Sort the input array in non-decreasing order. This enables the two-pointer technique and makes duplicate skipping straightforward.

2. Iterate with index i

Loop over each element as the first element of the triplet. Skip duplicates for i by checking if nums[i] == nums[i-1] and i > 0.

3. Two-pointer search

For each i, set left = i+1 and right = n-1. While left < right, compute sum = nums[i] + nums[left] + nums[right]. If sum < 0, increment left; if sum > 0, decrement right; if sum == 0, record triplet and skip duplicates for left and right.

4. Duplicate skipping logic

After finding a triplet, move left past all equal elements and right past all equal elements to avoid duplicate triplets. Also, skip duplicate i values at the start of the loop.

5. Prove correctness on adversarial inputs

For an array of all zeros, show that the algorithm finds exactly one triplet [0,0,0] by skipping duplicates at i, left, and right. Argue that any duplicate triplet would require equal elements at different positions, which are skipped.

Key Points to Mention

  • Time complexity: O(n^2) due to nested loops (sorting is O(n log n)).
  • Space complexity: O(1) extra space if output not counted, or O(n) for sorting depending on implementation.
  • Duplicate skipping for i: if i > 0 and nums[i] == nums[i-1], continue.
  • Duplicate skipping for left and right: while left < right and nums[left] == nums[left+1], left++; similarly for right.
  • Proof of correctness: invariant that after skipping, the next element is strictly greater, so no duplicate triplets.
  • Adversarial input: all zeros yields exactly one triplet [0,0,0] because duplicates are skipped.

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

Q2

Generalize the triplet sum problem to an arbitrary target T and then to k-sum for k >= 4 with pruning. What time complexity do you achieve, and how do you prevent integer overflow when summing large values?

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

Generalizing to T is trivial, just subtract from the target as you fix each element.

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AI HintsAI Generated

Suggested Approach

Start by generalizing the classic 3-sum problem to an arbitrary target T, then extend to k-sum using recursion with sorting and pruning. Discuss the time complexity trade-offs and how to prevent integer overflow using safe arithmetic or wider data types.

Pro tip: Emphasize that pruning is key to reducing the search space in practice, and mention that using 64-bit integers or Python's arbitrary precision can prevent overflow, but be mindful of performance implications.

1. Generalize to arbitrary target T

Explain how to modify the two-pointer approach for 3-sum to work with any target T by adjusting the sum comparison.

2. Extend to k-sum recursively

Describe a recursive function that reduces k-sum to (k-1)-sum by fixing one element and recursively solving for the remaining sum.

3. Incorporate pruning

Discuss pruning techniques such as early termination when the smallest or largest possible sum exceeds the target, and skipping duplicates.

4. Analyze time complexity

State that the time complexity is O(n^{k-1}) for the recursive approach, and explain how pruning can improve average-case performance.

5. Address integer overflow

Mention using 64-bit integers (long long in C++), Python's arbitrary precision, or checking for overflow before addition.

Key Points to Mention

  • Two-pointer technique for 2-sum and its extension to 3-sum with arbitrary target.
  • Recursive reduction from k-sum to (k-1)-sum.
  • Pruning strategies: sorting, early break, and duplicate skipping.
  • Time complexity: O(n^{k-1}) with potential improvements via pruning.
  • Integer overflow prevention: use of 64-bit integers, arbitrary precision, or safe addition checks.
  • Trade-offs between pruning effectiveness and overhead.

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

Q3

If the array has 5 million elements and memory is capped at 512 MB, how would you approach finding all triplets using an external-memory strategy? What is the I/O complexity? And if exact enumeration is infeasible, how would you approximate the count of unique triplets using probabilistic sketches?

System DesignAlgorithms & Data Structures
Author's notes

This part blindsided me.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem constraints and assumptions, then outline an external-memory algorithm that sorts the array and uses a streaming approach to find triplets, analyzing I/O complexity. If exact enumeration is infeasible, propose a probabilistic sketch like HyperLogLog to estimate the number of unique triplets, discussing trade-offs.

Pro tip: Emphasize that the choice of algorithm depends on the triplet definition (e.g., sum to zero, distinct values) and that you would validate the probabilistic estimate with a small-scale exact computation.

1. Clarify the problem

Ask clarifying questions about the triplet definition (e.g., sum to zero, distinct elements) and whether the array fits in memory. Confirm that 5 million elements (e.g., 8 bytes each) exceed 512 MB, necessitating external memory.

2. Design external-memory algorithm

Propose sorting the array using external merge sort (O(N log N) I/O), then for each element, use two pointers to find pairs that form a triplet, streaming through the sorted data. Alternatively, use a hash-based approach with partitioning.

3. Analyze I/O complexity

State that external sorting takes O(N log_{M/B} (N/B)) I/Os, and the two-pointer scan adds O(N) I/Os, so overall I/O complexity is dominated by sorting. Mention that if the array is already sorted, it's O(N) I/Os.

4. Propose probabilistic approximation

If exact enumeration is infeasible, suggest using a probabilistic sketch like HyperLogLog to estimate the number of unique triplets. Describe how to hash each triplet and feed into the sketch, noting that this requires generating triplets on the fly.

5. Discuss trade-offs and validation

Compare exact vs. approximate methods in terms of time, space, and accuracy. Mention that the sketch provides an estimate with a small relative error, and suggest validating on a smaller dataset.

Key Points to Mention

  • External merge sort and its I/O complexity
  • Two-pointer technique for finding triplets in sorted data
  • HyperLogLog or similar sketches for cardinality estimation
  • Memory constraints and block size considerations
  • Trade-offs between exact and approximate methods
  • Validation of probabilistic estimates

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

Q4

Given a frequency map of values rather than the raw array, compute the number of unique triplets summing to a target T in O(m^2) time where m is the number of distinct values. Handle edge cases where two or all three values in a triplet are the same.

Algorithms & Data StructuresData Modeling
Author's notes

Favorite part of the whole interview, weirdly.

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AI HintsAI Generated

Suggested Approach

Iterate over pairs of distinct values (including the same value twice) from the frequency map, compute the required third value, and check its frequency to count valid triplets. Carefully handle cases where the third value equals one or both of the pair values by adjusting counts based on available frequencies.

Pro tip: Clarify whether triplets are counted by distinct value combinations or by multiplicities (i.e., number of index triplets). The frequency map approach naturally counts combinations of values; if multiplicities are needed, multiply frequencies appropriately.

1. Understand the problem and edge cases

Confirm that triplets are unordered and that values can repeat. Identify the three cases: all three distinct, two equal, and all three equal.

2. Iterate over pairs of distinct values

Use two nested loops over the distinct values (i from 0 to m-1, j from i to m-1) to consider all unordered pairs, including the same value twice.

3. Compute the third value and check frequency

For each pair (a, b), compute c = T - a - b. If c is in the frequency map, determine how many valid triplets can be formed based on the relative order and equality of a, b, and c.

4. Count triplets with careful frequency adjustments

For each case (all distinct, two equal, all equal), calculate the number of combinations using frequencies, ensuring no overcounting and that enough occurrences exist.

5. Sum and return the total count

Accumulate the counts from all valid pairs and return the total number of unique triplets.

Key Points to Mention

  • Time complexity O(m^2) due to nested loops over distinct values, and space complexity O(m) for the frequency map.
  • Handling of duplicate values: when a == b, need freq[a] >= 2; when a == c or b == c, need freq[c] >= 2; when all equal, need freq[a] >= 3.
  • Avoiding double-counting by enforcing an ordering (e.g., a <= b <= c) or by iterating pairs with i <= j and then ensuring c >= b.
  • Use of combinations formula: for distinct values, multiply frequencies; for two equal, use C(freq, 2) * freq(other); for all equal, use C(freq, 3).
  • Edge cases: target T may be achieved with values not present; frequencies may be zero; m may be small (e.g., m < 3).
  • Potential optimization: break early if a > T/3 or if b > (T - a)/2 to reduce iterations.

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