Start by clarifying the problem constraints (e.g., array size, duplicates, k validity) and then present a progression of solutions: sorting (O(n log n)), min-heap of size k (O(n log k)), and Quickselect (average O(n)). Emphasize that Quickselect is the most efficient for large arrays, but discuss trade-offs like worst-case O(n^2) and the need for randomization or median-of-medians to guarantee linear time.
Pro tip: Mention that in practice, for small k, a min-heap is often preferred due to its simplicity and guaranteed O(n log k) time, while Quickselect is better for large k or when average-case performance is acceptable. Also, note that Meta often values clean, bug-free code and clear communication over squeezing out the absolute best complexity.
Ask about input size, range of values, duplicates, and whether k is guaranteed valid. This shows attention to detail and helps choose the right approach.
Start with sorting the array (O(n log n)) and then improve to using a min-heap of size k (O(n log k)). Explain why these work and their trade-offs.
Explain the partition-based Quickselect algorithm that finds the k-th largest in average O(n) time. Describe how it avoids full sorting by recursively partitioning only the relevant side.
Acknowledge Quickselect's worst-case O(n^2) and mention randomization or median-of-medians to achieve guaranteed O(n). Discuss when to use each approach based on constraints.
Write clean code for the chosen approach, handling edge cases (k=1, k=n, duplicates). Walk through a small example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by clarifying the problem: confirm whether the tree is a binary search tree or a general binary tree, and whether nodes have parent pointers. For a general binary tree, use a recursive post-order traversal that returns the node if it matches either target, otherwise recurses into left and right subtrees; the first node where both sides return non-null is the LCA. Discuss time and space complexity, and mention iterative or parent-pointer alternatives if applicable.
Pro tip: Meta interviewers value clean, bug-free code and clear communication. Before coding, walk through a small example to validate your logic, and after coding, test edge cases like one node being an ancestor of the other or nodes not present in the tree.
Ask whether the tree is a BST or a general binary tree, whether nodes have parent pointers, and whether both nodes are guaranteed to be in the tree. This determines the optimal approach.
For a general binary tree without parent pointers, use a recursive post-order traversal. If parent pointers exist, you can find the intersection of paths to the root. For a BST, you can use the BST property to guide the search.
Describe the recursive function: if the current node is null or matches either target, return the current node. Recurse left and right; if both return non-null, the current node is the LCA; otherwise return the non-null child.
State that the time complexity is O(n) in the worst case, as each node is visited once, and space complexity is O(h) for the recursion stack, where h is the tree height.
Walk through a simple tree with nodes, including edge cases: one node is the ancestor of the other, nodes are in different subtrees, or one node is missing. Verify the algorithm returns the correct LCA.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by contrasting the general binary tree LCA solution (which requires traversing the entire tree) with the BST property that allows for a more efficient search. Explain that in a BST, you can find the LCA by traversing from the root and using the ordering property to decide which subtree to explore, achieving O(h) time and O(1) space. Emphasize the trade-offs and why this optimization is possible.
Pro tip: Mention that the BST property allows you to find the LCA without needing parent pointers or additional data structures, which is a common follow-up. Also, clarify that the O(h) time is optimal for a single query, but if multiple queries are expected, preprocessing (like Euler tour + RMQ) might be beneficial.
Briefly explain that LCA in a general binary tree typically requires a recursive traversal that checks both subtrees, resulting in O(n) time. This sets the baseline for comparison.
Explain that in a BST, for any node, all values in the left subtree are smaller and all in the right are larger. This ordering allows us to determine the LCA by comparing node values with the two target nodes.
Start from the root and traverse down: if both targets are smaller than the current node, move left; if both are larger, move right; otherwise, the current node is the LCA (since the targets split or one equals the current).
State that the time complexity is O(h) where h is the height of the tree (O(log n) for balanced BST, O(n) worst-case), and space is O(1) for iterative. Contrast with O(n) time and O(h) space for general binary tree.
Mention handling of duplicate values (if allowed), one node being ancestor of the other, and potential follow-ups like handling multiple queries or if the tree is not balanced.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Since each node has a parent pointer, we can treat the problem as finding the intersection of two linked lists (the paths from each node to the root). Use a two-pointer technique: advance both pointers one step at a time, and when one reaches the root, redirect it to the other node's starting point. They will meet at the LCA after at most two passes.
Pro tip: Mention that this approach uses O(1) extra space and runs in O(h) time, where h is the height of the tree, which is optimal. Also, note that if the nodes are not in the same tree, the algorithm will still terminate but return null, so you should handle that edge case.
Confirm that the tree is binary (or not), that nodes have parent pointers, and that the two nodes are guaranteed to be in the same tree. Ask if the LCA of a node with itself is the node itself.
Describe how to use two pointers starting at the given nodes. Move both one step up at a time. When a pointer reaches the root, redirect it to the other node's starting point. They will meet at the LCA.
Pick a small tree and trace the pointers to show how they meet at the LCA. This demonstrates understanding and helps catch off-by-one errors.
State that time complexity is O(h) where h is the height of the tree, and space complexity is O(1). Compare with alternative approaches like using a hash set of ancestors (O(h) space).
Discuss cases where one node is an ancestor of the other, nodes are in different trees, or the tree is skewed. Explain how the algorithm handles them.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Start by clarifying the constraints: what 'too large to fit in memory' means (e.g., billions of nodes, disk-resident, distributed) and whether the tree is static or dynamic. Then propose a disk-based or distributed algorithm that minimizes random I/O and memory usage, such as external memory BFS/DFS with parent pointers or a MapReduce-based approach.
Pro tip: Emphasize that you would first check if the tree is static and if parent pointers exist; if so, you can use a two-pointer technique with O(1) memory by traversing from the nodes to the root, which is far simpler than distributed processing.
Ask about tree size, memory limits, disk vs. distributed storage, static vs. dynamic, and whether parent pointers are available. This determines the feasible approaches.
If parent pointers exist, use a two-pointer technique (like intersection of linked lists) with O(1) memory. Otherwise, consider external memory BFS/DFS that processes nodes in blocks.
For disk-based, use blocked BFS with parent tracking stored on disk. For distributed, use MapReduce: map each node to its parent, then iteratively reduce to find ancestors of both nodes until they meet.
Discuss the number of passes over data, random vs. sequential I/O, and network shuffles in distributed setting. Optimize by sorting or partitioning to reduce costs.
Consider skewed trees, caching frequently accessed nodes, and using Bloom filters to avoid unnecessary disk reads. Also discuss trade-offs between time and space.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.