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

IntermediatePrefer not to say

Jun 2026

Summary

Google SWE coding round, one question the whole time, interval scheduling. Spent a decent chunk of the session just clarifying the problem before writing a single line, which felt risky but probably saved me from going down the wrong path.

Questions Asked (1)

Given a set of meetings and a number of available rooms, assign each meeting to a room such that the room with the smallest index is always preferred when multiple rooms are free. Return the room that held the most meetings.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The two-heap setup is the move here: one min-heap tracking free rooms by index, another tracking busy rooms by end time.

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AI HintsAI Generated

Suggested Approach

Clarify that meetings are given as intervals and that we need to simulate assigning each meeting to the lowest-index available room, then count meetings per room. Use a min-heap to track room availability by end time, and a separate min-heap for free room indices to always pick the smallest index. After processing all meetings, return the room with the maximum count (tie-break by smallest index).

Pro tip: Mention that the problem is essentially a greedy interval scheduling with resource allocation, and that using two heaps (one for busy rooms by end time, one for free room indices) yields an O(n log n) solution. Also note that if meetings are not sorted by start time, sort them first; if they are, skip sorting.

1. Clarify Inputs and Assumptions

Confirm that meetings are given as [start, end] intervals, rooms are indexed 0 to k-1, and that meetings are non-overlapping in time? Actually, they may overlap. Ask if meetings are sorted by start time. Clarify tie-breaking: smallest room index when multiple free.

2. Choose Data Structures

Use a min-heap for free room indices (initialized with all rooms) and a min-heap for busy rooms keyed by end time. Also maintain an array to count meetings per room.

3. Process Meetings in Order

Sort meetings by start time if needed. For each meeting, release all busy rooms whose end time <= current start time (move them to free heap). Then assign the meeting to the smallest free room index, increment its count, and push the room into the busy heap with its end time.

4. Find the Room with Most Meetings

After processing all meetings, scan the count array to find the room with the maximum count. If there's a tie, return the smallest index.

5. Analyze Complexity and Edge Cases

State time complexity O(n log n) due to sorting and heap operations, space O(n + k). Discuss edge cases: no meetings, more rooms than meetings, all meetings at same time, etc.

Key Points to Mention

Greedy assignment: always pick the smallest available room index.
Use of min-heap for free rooms to efficiently get the smallest index.
Use of min-heap for busy rooms to release rooms whose end time <= current start time.
Sorting meetings by start time if not already sorted.
Maintaining a count array to track meetings per room.
Time complexity O(n log n) and space O(n + k), where n is number of meetings and k is number of rooms.

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