Classic problem but the O(1) constraint is where people trip up.
Use a two-step in-place transformation: first transpose the matrix (swap elements across the main diagonal), then reverse each row. This achieves the 90-degree clockwise rotation with O(1) extra space and O(n^2) time. Clearly explain the mapping of each element and justify why the space is constant.
Pro tip: Mention that the rotation can be done in a single pass by swapping four elements at a time in a cycle, but the transpose+reverse method is simpler and less error-prone. Also, note that the problem assumes a square matrix; for non-square, the approach differs.
Confirm that the matrix is n×n, rotation is 90 degrees clockwise, and it must be done in-place with O(1) extra space. Ask if n can be 0 or 1, and if the matrix is mutable.
Describe how transposing swaps matrix[i][j] with matrix[j][i] for i < j, then reversing each row yields the clockwise rotation. Walk through a small example (e.g., 3×3) to illustrate.
State that transposing visits each element once (O(n^2)), and reversing rows also takes O(n^2), so overall time is O(n^2). Space is O(1) because only a temporary variable is used for swaps.
Cover n=0 (empty matrix), n=1 (no change), and large n (performance). Also mention that the algorithm works for any integer values, including negatives.
Briefly mention the four-way swap cycle method, which rotates four elements at a time in a single pass, but note it's more complex to implement correctly.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
The base LRU with a hashmap plus doubly linked list is something I could draw in my sleep, but the concurrency and TTL parts are where the conversation got interesting.
Start by outlining the core data structures: a hash map for O(1) lookups and a doubly linked list for O(1) eviction/insertion. Then discuss concurrency strategies (e.g., sharding, locks, or lock-free approaches) and finally explain how to integrate TTL using timestamps and lazy or active expiration.
Pro tip: Emphasize trade-offs: for example, sharding reduces lock contention but complicates global LRU ordering; TTL adds overhead but can be optimized with a min-heap or timing wheel. Apple values practical, scalable solutions that balance performance and complexity.
Ask about expected throughput, read/write ratio, consistency needs, and whether TTL is mandatory or optional. This shows you consider real-world usage before diving into design.
Explain using a hash map (key -> node) and a doubly linked list (nodes in access order). Describe how get moves a node to the front and put inserts/updates and evicts the tail when capacity is exceeded.
Discuss options: a single mutex (simple but contended), fine-grained locking (e.g., per-bucket locks with a global LRU list), or sharding the cache into independent segments. Mention trade-offs between simplicity and scalability.
Propose storing an expiration timestamp in each node. For expiration, use lazy deletion on access and/or a background thread with a min-heap or timing wheel to proactively remove expired entries. Discuss overhead and trade-offs.
Recap the design, highlighting how it meets O(1) average time, handles concurrency, and supports TTL. Mention potential optimizations and limitations (e.g., memory overhead, lock contention).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use two heaps: a max-heap for the lower half and a min-heap for the upper half, keeping their sizes balanced. Insertion involves adding to the appropriate heap and rebalancing, while median retrieval is O(1) by peeking at the heap tops. For odd total count, the median is the top of the larger heap; for even, it's the average of both tops.
Pro tip: Mention that this approach also works for streaming data with limited memory, and discuss how to handle duplicates and negative numbers gracefully.
Confirm that the data structure should support dynamic insertion and median queries, and discuss expected input size and data types.
Explain that a max-heap stores the smaller half and a min-heap stores the larger half, maintaining size balance.
Describe adding the new element to one heap based on comparison with the other heap's top, then rebalancing sizes so they differ by at most one.
For odd total, return the top of the larger heap; for even, return the average of the two heap tops.
State that insertion is O(log n) due to heap operations, median retrieval is O(1), and discuss handling empty stream or single element.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
For directed graphs I went with DFS and a three-color visited set to distinguish nodes in the current recursion stack from fully processed ones.
Start by clarifying the graph type (directed vs. undirected) and whether you need to return an example cycle. For directed graphs, use DFS with a recursion stack (or three-color marking) to detect back edges; for undirected graphs, use DFS with parent tracking or Union-Find, noting that each edge appears twice. If returning a cycle, maintain a parent map to reconstruct the path when a back edge is found.
Pro tip: Mention that for directed graphs, a simple visited set is insufficient—you must track nodes in the current recursion stack (or use colors) to distinguish cross edges from back edges. Also, note that Union-Find is often preferred for undirected graphs due to its near-linear time and simplicity, but DFS is needed if you must return the actual cycle.
Ask whether the graph is directed or undirected, if it may be disconnected, and whether you need to return an example cycle or just a boolean. This determines the algorithm and data structures.
For directed graphs, use DFS with a recursion stack (or three-color marking) to detect back edges. For undirected graphs, use DFS with parent tracking or Union-Find, being careful to ignore the edge back to the parent.
Write the DFS (or Union-Find) code, handling disconnected components by iterating over all nodes. For directed graphs, maintain a 'visiting' set (or color array) and a 'visited' set; for undirected, pass the parent to avoid false positives.
If required, maintain a parent map during DFS. When a back edge is found, trace back from the current node to the ancestor using the parent map to build the cycle path.
State that both approaches run in O(V+E) time and O(V) space. Discuss trade-offs: DFS is simpler for directed graphs and can return a cycle; Union-Find is often faster for undirected graphs but doesn't easily return the cycle.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.