The 'count each dasher once' constraint is what makes this non-trivial.
Use a sweep line algorithm over the start and end times, but since each dasher must be counted at most once, track distinct dasher IDs at each event. Process events in chronological order, maintaining a set of active dashers and updating the maximum size of that set.
Pro tip: Clarify the exclusive end time semantics: an end event at time t means the dasher is no longer busy at t, so process end events before start events at the same timestamp. Also, consider using a hash set for O(1) add/remove and a counter for the current number of distinct dashers.
Confirm that end times are exclusive, meaning a dasher is busy from startTime inclusive to endTime exclusive. Discuss edge cases like zero-duration intervals, multiple intervals for the same dasher, and simultaneous events.
Create events for each interval: (startTime, 'start', dasherId) and (endTime, 'end', dasherId). Sort events by time, and for ties, process 'end' events before 'start' events to respect exclusive end times.
Use a set to store dasher IDs currently busy. For each event: if 'start', add dasherId to the set; if 'end', remove dasherId. After each event, update the maximum size of the set.
The algorithm runs in O(N log N) time due to sorting, where N is the number of intervals. Space is O(N) for events and O(D) for the set, where D is the number of distinct dashers. Mention that this is optimal for comparison-based sorting.
Walk through a simple example, such as intervals (1,3) and (2,4) for the same dasher, to show that the maximum is 1, not 2. Also test with different dashers to ensure the count increases correctly.
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Start by clarifying the problem and defining the events (start and end times for each dasher), then outline the sweep-line algorithm with sorting and tie-breaking rules. Explain how to deduplicate simultaneous starts for the same dasher and ensure ends are processed before starts at the same time to maintain correctness.
Pro tip: Emphasize the importance of tie-breaking and deduplication for correctness, and mention that using a stable sort or a custom comparator ensures deterministic behavior. Also, discuss how this approach generalizes to similar interval problems.
Restate the problem to ensure understanding, and define what constitutes a start and end event for each dasher. Specify that each event has a time, type (start/end), and dasher ID.
For each dasher, create start and end events. Deduplicate multiple simultaneous starts for the same dasher by keeping only one start event per dasher per time, or by merging them into a single event.
Sort events by time. For ties, process end events before start events. If multiple starts for the same dasher at the same time, ensure they are deduplicated. Use a stable sort or custom comparator to enforce order.
Iterate through sorted events, updating the active set of dashers. For end events, remove the dasher; for start events, add the dasher. Track any required metrics (e.g., maximum concurrent dashers).
Explain that sorting takes O(n log n) and sweeping takes O(n), so overall O(n log n). Discuss edge cases like simultaneous starts/ends and deduplication impact.
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Sorting by dasherId first means events get grouped by dasher rather than by time, which completely breaks the sweep.
First, explain that sorting by (dasherId, time, delta) groups events by dasher but orders by time before delta, which can misorder events that occur at the same timestamp. Then, propose the correct sort key as (dasherId, time, delta) is actually correct if delta represents a sequence number or logical clock; otherwise, if delta is a duration, the correct key is (dasherId, time) with a stable sort or (dasherId, time, eventId) to preserve order. Clarify the meaning of delta and emphasize the need for a deterministic tie-breaker.
Pro tip: Mention that in distributed systems, relying solely on timestamps is risky due to clock skew; using a logical sequence number (like delta) as a tie-breaker is a common pattern, but it must be monotonic per dasher.
Break down the tuple: dasherId (grouping), time (primary sort), delta (secondary sort). Explain what each represents.
Sorting by (dasherId, time, delta) orders by time first, then delta. If delta is a duration or non-monotonic, events at the same time may be misordered.
If delta is a sequence number, (dasherId, time, delta) is correct. If delta is a duration, use (dasherId, time, eventId) or stable sort by (dasherId, time).
Provide a concrete example where the wrong sort leads to incorrect event ordering, and how the correct key fixes it.
Mention scalability, memory, and whether additional fields (like eventId) are available; consider using a composite key with a unique identifier.
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First, clarify the original problem: likely given a list of orders with start and end times, find the maximum number of overlapping orders (allowing overlaps). Then, for the variant where dashers cannot take overlapping orders, the problem becomes finding the maximum number of non-overlapping orders (i.e., maximum independent set of intervals). Explain that the peak concurrent orders is solved by sorting events and sweeping, while the non-overlapping variant is solved by greedy interval scheduling. Compare time and space complexities for both.
Pro tip: Mention that the non-overlapping variant is equivalent to the classic interval scheduling maximization problem, which can be solved greedily by sorting by end time. This shows you recognize the underlying algorithmic pattern and can connect it to known solutions.
Restate the original problem: given a list of orders with start and end times, find the maximum number of concurrent orders. Confirm that overlapping is allowed. Then restate the variant: dashers cannot take overlapping orders, so we need the maximum number of non-overlapping orders (i.e., maximum set of orders that can be assigned to a single dasher without conflicts).
Sort all start and end events by time. Sweep through events, incrementing a counter for starts and decrementing for ends, tracking the maximum. Time complexity: O(n log n) due to sorting; space: O(n) for events or O(1) extra if sorting in place.
This is the interval scheduling maximization problem. Sort intervals by end time, then greedily select the interval with the earliest end time that starts after the last selected end. Time complexity: O(n log n) for sorting; space: O(1) extra if sorting in place, or O(n) if storing selected intervals.
Both variants have O(n log n) time due to sorting. Space: peak concurrent can be O(n) for events array, but can be O(1) extra if using in-place sort and two pointers; non-overlapping is O(1) extra if only counting, or O(k) for selected intervals. Emphasize that the dominant factor is sorting.
Mention that if orders are already sorted, time can be O(n). Discuss edge cases: empty input, all overlapping, none overlapping. Also note that the non-overlapping variant assumes each dasher can take multiple orders sequentially, but the question might imply a single dasher? Clarify if needed.
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