Meta·Software Engineer·Technical Phone Screen·Senior

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Jun 2026Remote

Summary

Meta SWE coding round with two back-to-back algorithm problems. The first was a tree problem with a space constraint that tripped me up, and the second escalated from a basic sort to heap and quickselect territory pretty fast.

Questions Asked (2)

Given two nodes in a rooted tree where each node has a parent pointer, implement a function to find their lowest common ancestor in O(h) time and O(1) extra space, without modifying the nodes.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The O(1) space constraint is what makes this annoying.

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AI HintsAI Generated

Suggested Approach

First, clarify the constraints and edge cases, then explain that since we cannot modify nodes or use extra space, we must compute the depths of both nodes by traversing to the root, then align the deeper node by moving up the difference, and finally move both pointers up in tandem until they meet. Emphasize that this achieves O(h) time and O(1) space.

Pro tip: Mention that if parent pointers are not available, the problem becomes harder and may require extra space, but since they are given, we can avoid storing paths. Also, note that the O(h) time is optimal because in the worst case we may need to traverse the height of the tree.

1. Clarify assumptions and edge cases

Confirm that the tree is rooted, nodes have parent pointers, and we cannot modify nodes or use extra space. Discuss edge cases: one node is ancestor of the other, nodes are the same, tree is skewed, etc.

2. Compute depths of both nodes

Traverse from each node up to the root to determine its depth. This takes O(h) time and O(1) space since we only keep counters.

3. Align depths

Move the deeper node up by the difference in depths so that both pointers are at the same depth.

4. Find LCA by moving up in tandem

While the two pointers are not equal, move both up one step at a time. When they become equal, that node is the LCA.

5. Analyze complexity and trade-offs

State that time complexity is O(h) and space is O(1). Discuss why this is optimal and mention alternative approaches (e.g., using hash sets) and their trade-offs.

Key Points to Mention

Depth calculation by traversing to root using parent pointers.
Aligning depths by moving the deeper node up.
Simultaneous upward traversal until pointers meet.
Time complexity O(h) where h is the height of the tree.
Space complexity O(1) because only a few pointers/counters are used.
No modification of nodes and no extra data structures.
Handling edge cases: same node, one ancestor of the other, skewed tree.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.

Given an array of n 2D points and an integer k, return the k points closest to the origin. Start with a naive sort-based solution, then implement a more efficient approach using either a max-heap or quickselect. Also define the minimal heap interface your solution relies on.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The naive sort part was basically a warmup, took two minutes.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem and constraints, then present a naive O(n log n) sort-based solution. Next, implement a more efficient O(n log k) max-heap solution, and optionally discuss quickselect for average O(n). Finally, define the minimal heap interface your solution relies on.

Pro tip: Mention that you avoid computing square roots by comparing squared distances, and discuss trade-offs between heap and quickselect based on k and n.

1. Clarify requirements and constraints

Ask about input size, whether k is guaranteed valid, if points can be duplicated, and if the output order matters. This shows attention to detail.

2. Naive sort-based solution

Compute squared distances for all points, sort them, and return the first k. Analyze time O(n log n) and space O(n).

3. Efficient max-heap solution

Iterate through points, maintain a max-heap of size k based on distance. For each point, if heap size < k, push; else if distance < heap top, pop and push. Finally, extract all elements. Time O(n log k), space O(k).

4. Alternative quickselect approach

Use quickselect to partition points by distance until the k-th element is in place, then return the first k. Average time O(n), worst O(n^2). Discuss randomized pivot to avoid worst-case.

5. Define minimal heap interface

Specify operations needed: push (insert), pop (remove max), top (peek max), size, and empty. This abstraction allows using any heap implementation.

Key Points to Mention

Use squared Euclidean distance to avoid floating-point precision issues and unnecessary sqrt operations.
Time and space complexity analysis for each approach: sort O(n log n), heap O(n log k), quickselect average O(n).
Trade-offs: heap is better when k is small, quickselect is better when k is close to n, but quickselect has worst-case O(n^2).
Handling edge cases: k=0, k=n, empty array, duplicate points.
Heap interface: push, pop, top, size, empty; can be implemented with a binary heap or language-provided priority queue.
Potential follow-up: how to handle streaming data or if points are added dynamically.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.