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Airbnb·Machine Learning Engineer·Technical Phone Screen·Senior

Senior
Apr 2026

Summary

Airbnb ML Engineer interview with a coding round that was basically a coin change problem dressed up in airport logistics clothing. The theme was cute but the problem was pretty standard once you stripped the flavor text away.

Questions Asked (1)

Q1

Given a target amount and a list of canister capacities (each usable unlimited times), find the minimum number of canisters that sum to exactly the target. Return -1 if it's impossible.

Algorithms & Data Structures
Author's notes

Classic unbounded knapsack / coin change in disguise.

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AI HintsAI Generated

Suggested Approach

Recognize this as the classic coin change problem (minimum coins) and solve it using dynamic programming. Define dp[i] as the minimum number of canisters to make amount i, initialize dp[0]=0 and others to infinity, then iterate amounts from 1 to target, updating dp[i] = min(dp[i], dp[i - capacity] + 1) for each capacity. Return dp[target] if finite, else -1.

Pro tip: After presenting the DP solution, mention that BFS on the amount graph is an alternative with the same O(target * n) time but can be more intuitive for finding the minimum number of steps. Also, briefly discuss edge cases like target=0 (return 0) and capacities larger than target (ignore them).

1. Clarify the problem

Confirm that canisters can be reused unlimited times, order doesn't matter, and we need the exact target. Ask about constraints (e.g., target size, number of capacities) to choose the right approach.

2. Define the DP state

Let dp[i] be the minimum number of canisters needed to make amount i. Initialize dp[0] = 0 and dp[i] = infinity for i > 0.

3. Fill the DP table

For each amount i from 1 to target, iterate over each capacity c. If i >= c and dp[i - c] is not infinity, update dp[i] = min(dp[i], dp[i - c] + 1).

4. Return the result

After filling the table, if dp[target] is still infinity, return -1; otherwise return dp[target].

5. Analyze complexity and edge cases

State time complexity O(target * n) and space O(target). Mention edge cases: target=0 returns 0, capacities > target are ignored, and impossible cases return -1.

Key Points to Mention

  • Dynamic programming approach with state dp[i] = min canisters for amount i
  • Time and space complexity: O(target * n) time, O(target) space
  • Edge cases: target = 0, capacities larger than target, impossible target
  • Alternative approach: BFS on the amount graph (also O(target * n))
  • Optimization: iterate capacities in outer loop for unbounded knapsack variant
  • Handling large targets: consider if target is huge, DP may be infeasible; discuss greedy or other heuristics if applicable

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