Google·Software Engineer·Onsite - Coding / Algorithms·Intermediate

Intermediate

Jun 2026

Summary

Google SWE coding round, one problem the whole session. The question was a scheduling/heap problem and the interviewer wanted both a working solution and a complexity discussion. Not a bad experience but definitely not easy.

Questions Asked (1)

Given n rooms and a list of meetings with start and end times, assign each meeting to the lowest-index available room. If no room is free, the meeting waits until one opens and keeps its original duration. Return the room index that hosted the most meetings, breaking ties by lowest index. Walk through a two-heap solution and analyze correctness plus time and space complexity.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I knew heaps were the right move pretty quickly but fumbled on the tie-breaking part.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem constraints and edge cases, then explain the two-heap approach: sort meetings by start time, use a min-heap for available rooms and another min-heap for occupied rooms keyed by end time. Simulate the process, updating meeting counts, and finally return the room with the most meetings, breaking ties by lowest index.

Pro tip: Emphasize that the two-heap solution is optimal because it efficiently manages room availability in O(log n) time per operation, and mention that the tie-breaking rule requires careful tracking of room indices.

1. Clarify the problem

Ask about constraints: input format, whether meetings are sorted, if durations are fixed, and how ties are broken. Confirm that waiting meetings keep their original duration.

2. Outline the two-heap strategy

Explain that you'll sort meetings by start time, use a min-heap for available rooms (initialized with all room indices) and a min-heap for occupied rooms keyed by end time. This allows efficient assignment and release of rooms.

3. Simulate the process

Iterate through meetings: release rooms whose end time <= current start, assign the lowest-index available room (or wait if none), and push the meeting's end time to the occupied heap. Track meeting counts per room.

4. Handle waiting meetings

If no room is available, the meeting waits until the earliest room is free. Its start time becomes that room's end time, and its end time is start + original duration. Assign that room.

5. Analyze correctness and complexity

Argue that the greedy choice of lowest-index room is optimal for tie-breaking. Time complexity: O(m log m + m log n) due to sorting and heap operations. Space complexity: O(n + m) for heaps and counts.

Key Points to Mention

Sorting meetings by start time to process in chronological order.
Using two min-heaps: one for available rooms (by index) and one for occupied rooms (by end time).
Handling waiting meetings by updating start time to the earliest available room's end time.
Tracking meeting counts per room and breaking ties by lowest index.
Time complexity: O(m log m + m log n) where m is number of meetings and n is number of rooms.
Space complexity: O(n + m) for heaps and auxiliary data structures.

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