Salesforce·Backend Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

May 2026

Summary

Salesforce backend round, one algorithmic problem the whole time. Pretty standard greedy/heap territory but I fumbled the implementation details more than I'd like to admit.

Questions Asked (1)

Given a list of events each with a start and end day, find the maximum number of events you can attend if you can only attend one per day.

Algorithms & Data Structures

Author's notes

I knew the greedy angle pretty quickly: sort by start day, use a min-heap keyed by end day, iterate day by day and always pick the event expiring soonest.

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AI HintsAI Generated

Suggested Approach

Clarify the problem constraints and edge cases, then propose a greedy algorithm that sorts events by end day and selects the earliest-ending event for each available day. Explain how this maximizes the count and analyze the time complexity.

Pro tip: Mention that this is a classic interval scheduling problem and that the greedy choice of earliest end time is optimal; also discuss how to handle multiple events with the same end day by picking any.

1. Clarify the problem

Ask if events can span multiple days, if start and end are inclusive, and if there are constraints on the number of events or days. Confirm that you can attend at most one event per day and that events cannot be partially attended.

2. Identify the algorithmic pattern

Recognize this as an interval scheduling maximization problem. The optimal strategy is to sort events by their end day and greedily select events that start after the last attended day.

3. Design the greedy algorithm

Sort events by end day ascending. Initialize lastAttendedDay to -infinity and count to 0. Iterate through events: if event.start > lastAttendedDay, attend it, increment count, and set lastAttendedDay = event.end. Return count.

4. Analyze complexity and edge cases

Time complexity is O(n log n) due to sorting, space O(1) or O(n) depending on sort. Discuss edge cases: no events, all events on same day, events with same end day, and events that start and end on the same day.

5. Test with examples

Walk through a small example to verify correctness, e.g., events = [[1,2],[2,3],[3,4]] yields 3, while [[1,2],[1,2],[1,2]] yields 1. Explain why greedy works by exchange argument.

Key Points to Mention

Greedy algorithm: sort by end day and pick earliest ending event that doesn't conflict.
Proof of optimality via exchange argument: replacing any event with the earliest-ending one doesn't reduce the count.
Time complexity: O(n log n) for sorting, O(n) for iteration.
Space complexity: O(1) extra space if sorting in-place, otherwise O(n).
Handling multiple events with the same end day: any can be chosen without affecting optimality.
Edge cases: empty list, single event, all events overlapping, events with same start and end.

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