Classic interval overlap problem but reframed around riders and orders.
Clarify that this is the classic interval partitioning problem, then explain that the minimum number of riders equals the maximum number of overlapping orders at any point in time. Present an efficient O(n log n) sweep-line algorithm: sort all start and end events, then scan to track the current number of active orders and record the maximum.
Pro tip: Mention that this problem is equivalent to finding the chromatic number of an interval graph, which is solvable in polynomial time, and that the greedy sweep-line approach is optimal. Also, discuss how this applies to real-world resource allocation at DoorDash, such as dynamic rider scheduling.
Confirm that orders are intervals [start, end) and that a rider can handle only one order at a time. Ask if orders can be preempted or if there are other constraints.
Recognize that the minimum number of riders needed is exactly the maximum number of orders that overlap at any single point in time. This is a known result for interval graphs.
Use a sweep-line approach: create events for each start (+1) and end (-1), sort them by time (with ends before starts if intervals are half-open), then iterate to compute the running sum and track the maximum.
State that the algorithm runs in O(n log n) time due to sorting and O(n) space for events. Discuss edge cases like empty input, simultaneous start/end, and large inputs.
Relate the problem to resource allocation in delivery systems, and mention how ML can predict order volumes to optimize rider scheduling dynamically.
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