Classic meeting rooms problem but they pushed hard on the 'very large inputs' part.
Start by clarifying the interval semantics (closed-open) and edge cases, then present the sweep-line algorithm: create events for starts (+1) and ends (-1), sort them, and track the running count to find the maximum concurrent meetings. Follow with complexity analysis (O(n log n) time, O(n) space) and discuss scaling strategies like external sorting or streaming for very large inputs.
Pro tip: Emphasize that closed-open intervals mean a meeting ending at time t does not overlap with one starting at t; handle this by processing end events before start events at the same timestamp. This detail shows precision and avoids off-by-one errors.
Confirm that intervals are closed-open, discuss empty input, single meeting, and simultaneous start/end. Ask if the input fits in memory or if it's a stream.
Explain creating events: for each interval [start, end), add (start, +1) and (end, -1). Sort events by time, with end events before start events at the same time. Iterate, maintaining a running count and tracking the maximum.
Sorting takes O(n log n) time, and the sweep takes O(n) time, so overall O(n log n). Space is O(n) for the events array. Mention that this is optimal for comparison-based sorting.
If data doesn't fit in memory, use external sorting or a streaming approach with a min-heap of end times. Alternatively, if times are bounded, use counting sort or a difference array for O(n + T) time.
Mention the min-heap approach: sort intervals by start time, use a min-heap of end times; for each meeting, if it starts after the earliest end, reuse a room; else add a room. Compare with sweep-line in terms of simplicity and constant factors.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
BFS with a null-sentinel trick, once you hit a null node any subsequent non-null means it's not complete.
Start by clearly defining a complete binary tree: all levels are fully filled except possibly the last, which is filled from left to right. Then present a level-order (BFS) traversal that detects the first missing child and ensures no subsequent node has a child. Discuss time and space complexity, edge cases, and test cases.
Pro tip: Mention that a complete binary tree can be efficiently checked using an array representation with index validation, but the BFS approach is more intuitive and avoids potential integer overflow for large trees. Also, note that TikTok values clean, bug-free code and clear communication, so walk through your logic step-by-step.
State the definition of a complete binary tree and clarify that the tree may be empty or have only one node.
Propose a level-order traversal using a queue. Explain that you will track whether a node has been seen without a left or right child.
Describe the BFS process: enqueue root, then for each node, check its children. If a missing child is encountered, set a flag; if any node after that has a child, return false. Handle null children appropriately.
State that time complexity is O(n) since each node is visited once, and space complexity is O(n) in the worst case (e.g., a perfect binary tree) due to the queue.
List edge cases: empty tree, single node, tree with only left children, tree with a gap in the last level, and a full binary tree. Suggest test cases covering these scenarios.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.