← Voleon Interview Insights

Voleon·Machine Learning Engineer·Technical Phone Screen·Intermediate

Intermediate
May 2026

Summary

Voleon MLE interview mixed a research background chat with two coding problems back to back. Pretty standard algo fare but the combination with the research discussion made it feel a bit disjointed, like they weren't sure which direction they wanted to take it.

Questions Asked (2)

Q1

Given K eggs and N floors, what is the minimum number of moves needed to find the critical floor in the worst case?

Algorithms & Data Structures
Author's notes

This one hurt.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Start by clarifying the problem: we need the minimum number of trials in the worst case to determine the highest floor from which an egg can be dropped without breaking. Then present the dynamic programming recurrence and explain how to optimize it, possibly using binary search or mathematical insights. Finally, discuss time and space complexity and potential optimizations.

Pro tip: Mention that this problem is equivalent to finding the smallest number of trials t such that the sum of binomial coefficients C(t, i) for i=1..K is at least N, which gives an O(K log N) solution. This shows deep understanding and avoids the common O(KN^2) DP.

1. Clarify the problem

Confirm that we need the minimum number of moves in the worst case to find the critical floor, and that eggs can break or survive. Define the critical floor as the highest floor from which an egg does not break.

2. Define the DP state and recurrence

Let dp[k][n] be the minimum moves with k eggs and n floors. The recurrence is dp[k][n] = 1 + min_{x=1..n} max(dp[k-1][x-1], dp[k][n-x]), with base cases dp[1][n] = n and dp[k][0] = 0.

3. Optimize the DP

Explain that naive DP is O(K N^2) but can be optimized to O(K N log N) using binary search on x because the max function is monotonic. Alternatively, use the inverse DP: find the maximum floors coverable with k eggs and m moves.

4. Present the mathematical solution

Show that the answer is the smallest t such that sum_{i=1}^K C(t, i) >= N. This can be computed in O(K log N) by binary searching t and summing binomial coefficients.

5. Analyze complexity and edge cases

Discuss time and space complexity of the chosen approach, and handle edge cases like K=1, K>=log2(N), or N=0. Mention that if K is large enough, binary search on floors works.

Key Points to Mention

  • Dynamic programming recurrence and base cases
  • Optimization using binary search or inverse DP
  • Mathematical formulation with binomial coefficients
  • Time and space complexity trade-offs
  • Edge cases: K=1, K>=log2(N), N=0
  • Connection to classic egg dropping puzzle and its variants

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

Q2

Given an integer array, find all unique triplets that sum to zero.

Algorithms & Data Structures
Author's notes

Sort plus two pointers, pretty mechanical once you remember to skip duplicates.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Start by clarifying the problem constraints (e.g., array size, duplicates, expected time complexity). Then propose the optimal O(n^2) solution: sort the array and use a two-pointer technique for each fixed element, skipping duplicates to ensure uniqueness. Discuss trade-offs with brute force and hash-based approaches.

Pro tip: Mention that sorting enables efficient duplicate skipping and two-pointer search, and that the O(n^2) time complexity is optimal for this problem since the output can be O(n^2) in the worst case. Also, handle edge cases like arrays with fewer than 3 elements.

1. Clarify and Confirm

Ask about input size, duplicate handling, and expected time/space complexity. Confirm that triplets must be unique and indices cannot be reused.

2. Outline Approaches

Briefly describe brute force O(n^3), hash map O(n^2) with extra space, and the optimal sort + two-pointer O(n^2) with O(1) extra space (excluding output).

3. Detail Optimal Solution

Explain: sort the array; for each index i, skip duplicates; use two pointers left=i+1 and right=n-1 to find pairs summing to -nums[i]; skip duplicates for left and right.

4. Analyze Complexity

State time complexity O(n^2) due to nested loops, and space complexity O(1) extra (or O(n) if counting sorting space).

5. Test with Examples

Walk through a small example like [-1,0,1,2,-1,-4] to demonstrate correctness and duplicate handling.

Key Points to Mention

  • Sorting the array to enable two-pointer technique and duplicate skipping.
  • Two-pointer approach for each fixed element to find pairs summing to its negation.
  • Skipping duplicates for the fixed element and for left/right pointers to ensure unique triplets.
  • Time complexity O(n^2) and space complexity O(1) extra (or O(n) due to sorting).
  • Edge cases: array length < 3, all zeros, no valid triplets.
  • Comparison with hash map approach: O(n^2) time but O(n) space, and potential duplicate handling issues.

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