The large timestamp constraint is the whole point of the problem.
Use a sweep line algorithm: create events for interval starts (+1) and ends (-1), sort them by time, and sweep through to maintain a running count of active users. Output segments between consecutive event times where the count is non-zero, merging adjacent segments with the same count.
Pro tip: Clarify whether intervals are inclusive and how to handle zero-count gaps; explicitly state that you'll skip segments with count 0 and merge consecutive segments with equal counts to ensure disjointness.
Confirm if intervals are inclusive [start, end] and how to treat touching intervals (e.g., [1,2] and [2,3] overlap at 2). Discuss handling of zero-length intervals and large timestamps.
For each interval, create a start event (+1) at start time and an end event (-1) at end time (or end+1 if inclusive). Sort events by time, processing starts before ends at the same timestamp if intervals are inclusive.
Iterate through sorted events, maintaining a running count. At each event time, record the count before applying the event's delta, then apply the delta. This gives the count for the segment starting at that time.
For each consecutive pair of event times, if the count between them is >0, add a segment [time_i, time_{i+1}) with that count. Merge adjacent segments with the same count to ensure disjointness.
State time complexity O(n log n) due to sorting, space O(n). Walk through a small example to verify correctness, including edge cases like no overlap or all overlapping.
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This one surprised me more than it should have.
Start by checking necessary conditions: total cells must equal sum of counts, and each count must be at most half the total cells (or a tighter bound based on grid dimensions). Then use a constructive algorithm like spiral filling or row-by-row serpentine filling to assign cells to types, ensuring each type forms a single connected region. If no construction works, return impossible.
Pro tip: Mention that for a single connected region, the count of any type cannot exceed the number of cells in a spanning tree of the grid (R*C - (R-1)*(C-1) = R+C-1) if other types are present, but more practically, use a BFS/DFS to verify connectivity after construction and backtrack if needed. Also, note that Apple values clean, efficient code and edge-case handling.
Check that sum of counts equals R*C and that no count exceeds the maximum possible for a connected region (e.g., R*C - (number of other types) if others must be connected). Also consider parity or shape constraints.
Select a filling pattern that naturally creates connected regions, such as serpentine row-by-row filling, spiral filling, or using a priority queue to always extend the smallest region.
After filling, run BFS/DFS from one cell of each type to ensure all cells of that type are reachable. If not, adjust the filling order or backtrack.
If no valid grid is found after reasonable attempts, return impossible. Consider cases like single row/column, one type dominating, or counts that force disconnection.
Discuss time/space complexity of your approach (e.g., O(R*C) for filling and verification) and trade-offs between greedy construction and backtracking.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Pretty standard answer here: a sorted structure like a balanced BST or two heaps (one min, one max) with lazy deletion gets you O(log n) on everything.
Start by clarifying requirements: multiset means duplicates allowed, and operations should be efficient. Propose a balanced BST (e.g., Red-Black Tree) or two heaps with lazy deletion, discussing O(log n) for all operations. Then compare with a bucket/counting approach (e.g., array of counts) which gives O(1) for insert and O(1) amortized for getMin/getMax if the range is small, but O(range) for popMin/popMax in worst case, and discuss trade-offs based on data distribution and constraints.
Pro tip: Mention that in practice, for bounded integer ranges, a Fenwick tree over counts can support all operations in O(log U) where U is the universe size, and that Apple often values practical optimizations for constrained environments.
Ask about the expected number of operations, the range of integers, memory constraints, and whether the multiset is static or dynamic. This determines the best data structure.
Describe a balanced BST (e.g., Red-Black Tree) or a combination of a min-heap and max-heap with lazy deletion, achieving O(log n) for all operations.
For the balanced BST, all operations are O(log n) time and O(n) space. For two heaps with lazy deletion, insert is O(log n), getMin/getMax O(1), popMin/popMax O(log n) amortized.
If the integer range is small (e.g., 0 to 10^6), use an array of counts. Insert is O(1), getMin/getMax can be O(1) with maintained pointers, but popMin/popMax may require scanning, leading to O(range) worst-case.
Summarize when each approach is better: bucket/counting for small, dense ranges with frequent inserts and infrequent pops; balanced BST for large or sparse ranges with frequent pops.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.