My first instinct was just sort the planes once and for each order binary search for the eligible subset, then iterate through filling greedily.
Clarify the problem constraints and edge cases, then propose an efficient data structure solution that maintains planes sorted by departure time and supports range capacity queries and updates. Explain how to use a segment tree or Fenwick tree with binary search to achieve O((n + m) log n) time, and discuss trade-offs with simpler approaches.
Pro tip: Emphasize that the greedy allocation is optimal because filling earliest-departing planes first preserves later planes for future orders, and mention that using a segment tree with lazy propagation or a Fenwick tree with binary lifting can efficiently find the first plane with remaining capacity.
Ask about input sizes, whether orders are processed in timestamp order, and if planes can be partially filled. Confirm that ties in departure time are broken by index.
Propose maintaining planes in a segment tree keyed by departure time, storing remaining capacity. Support queries for total capacity after a timestamp and updates when capacity is consumed.
For each order, query total remaining capacity of planes departing after the order's timestamp. If insufficient, reject. Otherwise, greedily allocate by repeatedly finding the earliest-departing plane with remaining capacity and filling it until the order is satisfied.
Explain that each order requires O(log n) for the capacity query and O(k log n) for allocations, where k is the number of planes used. Since each plane can be filled at most once per order, total allocations across all orders is O(n + m), yielding O((n + m) log n).
Mention simpler O(nm) approaches and why they are inefficient. Compare segment tree vs. Fenwick tree with binary lifting, and note that a balanced BST with subtree sums could also work.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.