Amazon Software Engineer Interview Questions

I would first think about the recursive structure of the tree. For each node, I would search the left and right subtrees, and if both sides return a non-null result, then the current node is the lowest common ancestor. If only one side returns a value, I would propagate that value upward. The reason I choose recursion is that it matches the tree structure naturally and keeps the implementation clean. The time complexity is O(n) because I may visit every node, and the space complexity is O(h) for the recursion stack. I would also mention the BST-specific optimization if the interviewer asks for a simpler version.

I would use a queue because level-order traversal is naturally breadth-first. I would enqueue the root, then process nodes level by level by taking the current queue size at each iteration and collecting all nodes from that level before moving on. The reason I choose a queue is that it preserves insertion order, which matches the order I need for BFS. The time complexity is O(n) and the space complexity is O(n) in the worst case if the last level is very wide. I would also mention that if the interviewer wants zigzag traversal, I can adapt the same approach with a deque or level reversal.

I would first ask whether the graph is directed or undirected, because the solution changes. For a directed graph, I would use DFS with a recursion stack or colors to detect back edges. For an undirected graph, I would use DFS or Union-Find and make sure I do not treat the parent edge as a cycle. The reason I like DFS here is that it directly tracks the current path, which makes cycle detection straightforward. The time complexity is O(V + E), and the space complexity is O(V) for recursion or auxiliary state. I would also mention that if the interviewer wants topological sorting, cycle detection is part of that process.

I would first think about converting the tree into a string using preorder traversal with a special marker for null nodes. That makes the structure unambiguous when I reconstruct it later. On deserialization, I would read the tokens in the same order and rebuild the tree recursively. The reason I choose preorder with null markers is that it preserves structure cleanly and is easy to implement. The time complexity is O(n) for both serialization and deserialization, and the space complexity is O(n) because I store all nodes in the serialized form. I would also mention edge cases like an empty tree and skewed trees.

I would first think about the brute-force O(n^2) solution that checks every pair. To optimize it, I would use a hash map to store values I have already seen and look up the complement for the current number as I scan the array. That gives O(n) time and O(n) space. The reason I choose a hash map is that complement lookup becomes constant time, which is exactly what makes this solution efficient. I would also mention edge cases like duplicate numbers and ensuring I do not reuse the same index twice.