Airbnb·Software Engineer·Technical Phone Screen·Senior

SeniorPrefer not to say

Jul 2026Remote

Summary

Airbnb software engineering interview with a meaty algorithmic problem that required me to think through sorting, DP, and binary search all at once. Tougher than I expected for a single question.

Questions Asked (1)

Given a list of jobs each with a start time, end time, and reward value, find a subset of non-overlapping jobs that maximizes total reward. Return both the maximum reward and one valid set of job indices. Target O(n log n) time. Be prepared to explain your algorithm, prove it correct, analyze complexity, and walk through edge cases like zero-length jobs or identical timestamps.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I knew weighted job scheduling but blanked on how to reconstruct the actual job indices, not just the max value.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem and constraints, then propose a dynamic programming solution that sorts jobs by end time and uses binary search to find the latest non-overlapping job. Explain how to reconstruct the selected jobs by storing predecessor indices, and analyze time and space complexity to confirm O(n log n) time.

Pro tip: Emphasize that sorting by end time is crucial for the DP recurrence to work correctly, and proactively discuss how to handle edge cases like zero-length jobs or identical timestamps to show thoroughness.

1. Clarify requirements and constraints

Ask about input size, whether jobs can have zero duration, if timestamps are integers or floats, and whether any valid set is acceptable. Confirm that jobs are non-overlapping if one ends before the other starts.

2. Sort jobs by end time

Sort the jobs in ascending order of end time. This ordering ensures that when considering job i, all compatible jobs have indices less than i, enabling a clean DP recurrence.

3. Define DP state and recurrence

Let dp[i] be the maximum reward using jobs up to index i. For each job i, find the latest job j < i that does not overlap (end time ≤ start time of i) using binary search. Then dp[i] = max(dp[i-1], reward[i] + dp[j]).

4. Reconstruct the selected jobs

Maintain a parent array to record whether job i was included. After computing dp, backtrack from the last index to collect the indices of chosen jobs.

5. Analyze complexity and edge cases

State that sorting takes O(n log n), binary search for each job takes O(log n), and DP takes O(n), so overall O(n log n) time and O(n) space. Discuss handling zero-length jobs (treat as non-overlapping if end ≤ start) and identical timestamps (stable sort or tie-breaking).

Key Points to Mention

Sorting by end time is essential for the DP recurrence to consider all compatible jobs.
Binary search (or two-pointer) to find the latest non-overlapping job efficiently.
DP recurrence: dp[i] = max(dp[i-1], reward[i] + dp[latest_non_overlapping])
Reconstruction using a parent array to output one optimal set of job indices.
Time complexity O(n log n) due to sorting and binary search; space O(n).
Edge cases: zero-length jobs, identical start/end times, and empty input.

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