Implement node messaging and path discovery
Company: OpenAI
Role: Software Engineer
Category: Coding & Algorithms
Difficulty: medium
Interview Round: Technical Screen
Quick Answer: This question evaluates skills in distributed systems and algorithms, including reliable message-passing between adjacent nodes, graph traversal and path discovery, fault-tolerant protocol design, and message/time complexity analysis.
Part 1: Reliable Adjacent sendMessage
Constraints
- 0 <= max_retries <= 200000
- 0 <= len(events) <= 200000
- Each events[i] contains exactly two booleans
Examples
Input: ([(True, True)], 3)
Expected Output: (True, 1, 1)
Explanation: The message and ACK both succeed on the first attempt.
Input: ([(True, False), (True, True)], 3)
Expected Output: (True, 2, 1)
Explanation: The first ACK is lost, so the sender retries. The receiver sees a duplicate but delivers to the application only once.
Hints
- Track whether the receiver has already accepted the message once.
- The sender stops immediately on the first attempt where both the message reaches the receiver and the ACK reaches the sender.
Part 2: Count Total Nodes with Tree Message Passing
Constraints
- 0 <= n <= 200000
- If n > 0, edges has length n - 1
- 0 <= root < n when n > 0
- The input graph is a tree
Examples
Input: (0, [], -1)
Expected Output: (0, 0, 0)
Explanation: Edge case: empty network.
Input: (1, [], 0)
Expected Output: (1, 0, 0)
Explanation: A single-node tree needs no messages.
Hints
- In a tree protocol like this, each edge carries one request downward and one reply upward.
- The completion time in synchronous rounds is determined by the farthest leaf from the root.
Part 3a: Path from a Node to the Root
Constraints
- 0 <= len(parent) <= 200000
- If parent is non-empty, it represents a valid rooted tree
- If parent is empty, start will be -1
Examples
Input: ([-1, 0, 0, 1, 1], 4)
Expected Output: [4, 1, 0]
Explanation: Node 4 goes to 1, then to root 0.
Input: ([-1], 0)
Expected Output: [0]
Explanation: Edge case: a single-node tree.
Hints
- Keep following parent[start] until you reach -1.
- You do not need DFS or BFS because every node already knows its parent.
Part 3b: Path Between Two Arbitrary Nodes
Constraints
- 0 <= len(parent) <= 200000
- If parent is non-empty, it represents a valid rooted tree
- If parent is empty, u and v will both be -1
Examples
Input: ([-1, 0, 0, 1, 1, 2], 3, 5)
Expected Output: [3, 1, 0, 2, 5]
Explanation: The path goes up from 3 to the root and then down to 5.
Input: ([-1, 0, 1, 2], 3, 1)
Expected Output: [3, 2, 1]
Explanation: One node is an ancestor of the other.
Hints
- Build each node's path to the root, then find where those two paths stop matching.
- The last common node on the two root-paths is the lowest common ancestor.
Part 5: Simulate Failure Handling with Retries, ACKs, Idempotency, and Termination
Constraints
- 0 <= len(commands) <= 100000
- 0 <= len(events) <= 100000
- 0 <= max_retries <= 100000
- All command IDs are distinct
- Each ID in received_ids belongs to commands and is either the current command or an older command
Examples
Input: ([], [], 3)
Expected Output: (True, 0, [], 0)
Explanation: Edge case: there is nothing to send, so the protocol is already finished.
Input: ([7], [([7], False, False), ([7, 7], True, False)], 3)
Expected Output: (True, 1, [7], 2)
Explanation: The first ACK is lost, so the sender retries. The receiver applies command 7 only once.
Hints
- Keep separate state for what the sender is waiting for and what the receiver has already applied.
- A crash after processing received_ids still means the current command is unacknowledged unless the ACK was observed before the crash, which this model does not allow.