Start by clarifying requirements and constraints, then propose a hash map combined with a min-heap or balanced BST for efficient expiration. Discuss trade-offs between lazy and eager expiration, and how to maintain an accurate live key count.
Pro tip: Mention that using a min-heap for expiration can lead to stale entries; suggest lazy deletion with periodic cleanup or a timing wheel for better performance at scale. Also, highlight that the timestamp parameter allows deterministic testing and avoids reliance on system clocks.
Ask about expected scale, concurrency needs, and whether expired keys should be removed eagerly or lazily. Confirm that timestamps are provided for all operations.
Propose a hash map for O(1) key lookup and a min-heap or balanced BST for efficient expiration ordering. Discuss the trade-offs of each.
Explain how to check and remove expired entries during get/put operations (lazy) or via a background thread (eager). Ensure the live count is updated correctly.
Describe how to increment/decrement the count on insertions and deletions, and how to handle expired entries that are still in the data structure.
Mention alternatives like timing wheels for high throughput, and the impact of concurrency on design choices. Address time and space complexity.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
The O(n) constraint is what makes this non-trivial.
Use a single postorder traversal that returns both the height and balance status of each subtree, while collecting node values and depths in preorder. This avoids recomputing heights and achieves O(n) time. Then output the collected information in preorder.
Pro tip: Emphasize that you're combining two traversals: a postorder pass for height/balance and a preorder pass for output, but both can be done in one recursive function that processes nodes in postorder and stores results for later preorder output. This shows you understand traversal trade-offs and optimization.
Confirm that 'balanced' means the absolute difference between left and right subtree heights is at most 1 for every node. Ensure output order is preorder (root, left, right) and that O(n) time is required.
Create a helper function that takes a node and its depth, returns the height of the subtree rooted at that node, and also records the node's value, depth, and balance status. Use postorder to compute heights bottom-up.
In the helper, recursively get left and right subtree heights. The current node's height is 1 + max(leftHeight, rightHeight). It is balanced if the absolute difference <= 1 and both subtrees are balanced.
While recursing, store each node's value, depth, and balance status in a list in preorder (e.g., append before recursing). Alternatively, perform a separate preorder traversal after computing balance, but ensure it's O(n).
After the traversal, output the collected list in the required format. Discuss time and space complexity: O(n) time, O(h) space for recursion stack (or O(n) worst case).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.