Implement Caches, Undo, and Traversal
Company: Netflix
Role: Software Engineer
Category: Coding & Algorithms
Interview Round: Onsite
Quick Answer: This question evaluates proficiency with data structures, API design, and system-level concerns by combining timed in-memory caching, command undo/redo semantics, and depth-first traversal of a tree; it falls under the Coding & Algorithms domain and is commonly asked to assess practical implementation skills, correctness under edge cases, and the ability to reason about time and space complexity. The level of abstraction spans practical application—designing clean APIs and implementing core logic—while also requiring conceptual understanding of complexity analysis and background cleanup or state-management patterns.
Part 1: Timed Cache Simulation
Constraints
- 0 <= len(operations) <= 200000
- Keys are strings and values are integers
- 0 <= ttlSeconds <= 10^9
- 0 <= timestamp <= 10^9
- Operation timestamps are given in nondecreasing order
- Latest put for the same key overwrites the previous value
Examples
Input: [('put', 'a', 10, 5, 0), ('get', 'a', 4), ('cleanup', 5), ('get', 'a', 5)]
Expected Output: [10, 1, None]
Explanation: Key 'a' is valid at time 4, expires at time 5, is removed by cleanup at time 5, and is missing afterward.
Input: [('put', 'a', 1, 10, 0), ('put', 'a', 2, 3, 1), ('get', 'a', 3), ('cleanup', 5), ('get', 'a', 5), ('cleanup', 11), ('get', 'a', 11)]
Expected Output: [2, 1, None, 0, None]
Explanation: The second put overwrites the first. Cleanup removes only the current expired entry. The old heap record is stale and must be ignored.
Input: [('put', 'x', 7, 0, 10), ('cleanup', 10), ('get', 'x', 10)]
Expected Output: [1, None]
Explanation: A ttl of 0 means the entry expires immediately at the put timestamp.
Input: []
Expected Output: []
Explanation: No operations means no outputs.
Hints
- Use a hash map for fast key lookup on get.
- For background cleanup, keep expirations in a min-heap and use a version counter so stale heap entries from overwritten keys can be ignored.
Part 2: Undo Manager for Text Commands
Constraints
- 0 <= len(operations) <= 200000
- All typed strings are non-empty
- 1 <= k <= 200000 for delete operations
- Total length of all typed strings is at most 200000
- Operations are valid tuples in one of the listed formats
Examples
Input: [('type', 'ab'), ('type', 'cd'), ('delete', 3), ('undo',), ('undo',)]
Expected Output: ['ab', 'abcd', 'a', 'abcd', 'ab']
Explanation: Undo first restores the deleted text, then undoes the earlier type command.
Input: [('type', 'hi'), ('delete', 5), ('undo',), ('undo',)]
Expected Output: ['hi', 'hi', '', '']
Explanation: The delete fails because the buffer is too short, so it is not added to history. The first undo removes 'hi'. The second undo does nothing.
Input: [('undo',)]
Expected Output: ['']
Explanation: Undo on empty history is a no-op.
Input: [('type', 'a'), ('type', 'bc'), ('undo',), ('type', 'd'), ('delete', 1), ('undo',)]
Expected Output: ['a', 'abc', 'a', 'ad', 'a', 'ad']
Explanation: After undoing 'bc', typing 'd', deleting it, and undoing again, the buffer returns to 'ad'.
Hints
- Use a stack to store the inverse of each successful command.
- When deleting, save the removed substring so undo can restore it exactly.
Part 3: DFS Traversal of an Organization Tree
Constraints
- 0 <= number of nodes <= 100000
- Each node has an integer id
- Children are already listed in the required left-to-right visit order
- Node ids are unique
Examples
Input: (None,)
Expected Output: []
Explanation: An empty tree has no nodes to visit.
Input: ({'id': 1, 'children': []},)
Expected Output: [1]
Explanation: A single-node tree is visited by returning just that node's id.
Input: ({'id': 1, 'children': [{'id': 2, 'children': [{'id': 5, 'children': []}, {'id': 6, 'children': []}]}, {'id': 3, 'children': []}, {'id': 4, 'children': [{'id': 7, 'children': []}]}]},)
Expected Output: [1, 2, 5, 6, 3, 4, 7]
Explanation: Preorder visits the root first, then each subtree from left to right.
Input: ({'id': 10, 'children': [{'id': 20, 'children': [{'id': 30, 'children': [{'id': 40, 'children': []}]}]}]},)
Expected Output: [10, 20, 30, 40]
Explanation: In a chain-shaped tree, preorder follows the single path from top to bottom.
Input: ({'id': 0, 'children': [{'id': -1, 'children': [{'id': 2, 'children': []}, {'id': 3, 'children': []}]}, {'id': 5, 'children': []}]},)
Expected Output: [0, -1, 2, 3, 5]
Explanation: The traversal still follows preorder correctly even with negative ids.
Hints
- An iterative solution with an explicit stack avoids recursion depth issues.
- If using a stack, push children in reverse order so they are visited left to right.