Meta Software Engineer Interview Questions

I would first think about sorting all the points by distance and taking the first k, but that would do more work than needed. To optimize it, I would use a heap so I only keep the k closest points as I scan the list. Another option is quickselect if I want average O(n) time, but I would usually explain the heap first because it is easier to reason about in an interview. The main idea is that I do not need the entire ordering, only the best k elements. I would mention that the time complexity is O(n log k) with a heap and the space complexity is O(k). I would also point out edge cases like k equal to the number of points and duplicate distances.

Before designing it, I would clarify whether the goal is strict chronological order or a ranked feed based on relevance and engagement. At a high level, I would separate candidate generation from ranking, because I do not want to score every possible post for every user at request time. I would keep recent posts in a fast storage layer, generate a smaller set of candidate stories based on follows, interactions, and freshness, and then rank those candidates with a scoring model or rules. The trade-off is between relevance and latency, so I would balance personalization with a serving path that stays fast enough for real-time use. I would also think about caching, fan-out strategy, and failure fallback so the feed still works if ranking services are degraded.

I would use either BFS or DFS with two colors. As I traverse the graph, I color each node with one of two colors and make sure no edge connects two nodes of the same color. If I ever find a conflict, then the graph is not bipartite. The reason I choose this approach is that it is simple, efficient, and works in O(V + E) time. I would also mention that if the graph is disconnected, I need to run the traversal from every unvisited node. Edge cases include an empty graph and a graph with isolated vertices.

I would first think about using an extra array and copying elements into their new positions, which is the simplest way to reason about it. To optimize space, I would use the reverse trick: reverse the whole array, then reverse the first k elements, then reverse the remaining elements. The reason I choose this method is that it gives O(n) time and O(1) extra space while staying fairly easy to explain. I would also mention the direction of rotation because left and right rotation are mirror versions of the same idea. Edge cases include k larger than the array length, k equal to zero, and an empty array.

I would first count the frequency of each character using a hash map. Then I would scan the string again from left to right and return the first character whose count is one. The reason I choose this approach is that it is straightforward and runs in O(n) time. The space complexity is O(1) if the character set is fixed, or O(n) in the general case. I would also mention edge cases like an empty string and strings where every character repeats.