Google·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Google SWE coding round, one problem the whole time, heap-based room scheduling. Not a bad experience but I definitely underestimated how much they'd want to talk through complexity after the code was done.

Questions Asked (1)

You have n rooms (indexed 0 to n-1) and a list of meetings with start and end times. Meetings are processed in order of start time. Each meeting goes to the lowest-index free room; if none are free, it waits until the earliest room opens up and runs for its original duration. Return the room index that hosted the most meetings, with ties broken by lowest index. Walk through your implementation and complexity.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I knew pretty quickly it was a two-heap problem.

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AI HintsAI Generated

Suggested Approach

Clarify that meetings are assigned to the lowest-index free room, and if none are free, they wait until the earliest room becomes available. Use a min-heap to track room availability by end time and a counter array to tally meetings per room. After processing all meetings, return the room with the maximum count, breaking ties by lowest index.

Pro tip: Mention that the waiting behavior means a meeting's start time can be delayed, but its duration remains unchanged; this is crucial for correctly updating the room's next available time. Also, note that the room index tie-breaking is naturally handled by iterating from 0 to n-1 when finding the max.

1. Clarify and Restate

Confirm your understanding of the problem: meetings are sorted by start time, assigned to the lowest-index free room, and if none free, wait for the earliest room. Ask about edge cases like simultaneous meetings or empty input.

2. Choose Data Structures

Use a min-heap to track available rooms by their next free time, and a counter array to count meetings per room. The heap stores pairs (end_time, room_index) for rooms currently in use.

3. Process Meetings

Iterate through meetings in order. For each meeting, pop from the heap all rooms whose end_time <= meeting start, marking them free. If a free room exists, assign the lowest-index free room; otherwise, pop the earliest-ending room, update its end_time to meeting start + duration, and push it back. Increment the room's counter.

4. Find Result

After processing all meetings, scan the counter array from index 0 to n-1 to find the room with the maximum count, returning the first one encountered in case of ties.

5. Analyze Complexity

Explain that each meeting causes at most one heap push and pop, leading to O(m log n) time where m is the number of meetings and n is the number of rooms. Space is O(n) for the heap and counters.

Key Points to Mention

The need to sort meetings by start time (if not already sorted) or assume they are given sorted.
Using a min-heap keyed by end time to efficiently find the earliest available room.
Maintaining a separate min-heap or sorted structure for free rooms to quickly get the lowest-index free room.
Handling the case when no rooms are free: waiting until the earliest room opens, which may delay the meeting's start.
Tie-breaking rule: when multiple rooms have the same maximum meeting count, return the lowest index.
Time and space complexity analysis: O(m log n) time, O(n) space.

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