Interview questions · Bloomberg
Solve Magic Dictionary and unknown traversal
This interview question evaluates algorithm design, data structures, correctness, complexity, edge cases, and implementation details in a realistic interview setting. A strong answer for Solve Magic Dictionary and unknown traversal states assumptions, handles edge cases, explains trade-offs, and shows how to validate the result clearly.
LeetCode 676. Implement Magic Dictionary
Determine and implement traversal/encoding that outputs nodes as 'cia' for the given skewed binary tree and similar examples.
https://leetcode.com/problems/implement-magic-dictionary/description/
What a Strong Answer Covers
- A clear algorithm with the right data structures and enough pseudocode or code-level detail to implement it.
- A correctness argument that explains why the algorithm covers all required cases.
- Time and space complexity, plus at least one alternative approach when relevant.
- Focused tests for normal cases, edge cases, and failure modes.
Follow-up Questions
- How would the approach change if the input were streaming or too large for memory?
- What invariants would you assert in production code?
- Which tests would catch off-by-one, duplicate, or tie-breaking bugs?
Solution
Solution Alignment
The prompt asks for an implementation-level answer. The safest way to present it is to define the state, maintain clear invariants, then walk through complexity and tests.
Problem Restatement
Question LeetCode 676. Implement Magic Dictionary Determine and implement traversal/encoding that outputs nodes as 'cia' for the given skewed binary tree and similar examples. https://leetcode.com/problems/implement-magic-dictionary/description/
Recommended Approach
Choose traversal based on the required output. DFS is natural for subtree computations, reconstruction, and range pruning; BFS is natural for level order or side views. Keep per-depth or per-position state when the output depends on columns, rows, or depths.
Correctness
The implementation should maintain an invariant after each loop or operation that directly matches the problem statement. At termination, that invariant implies the returned value has considered every valid candidate exactly once, or has preserved the required data-structure state after every API call.
Complexity
Most tree traversals are O(n) time and O(h) recursion stack for DFS or O(w) queue space for BFS, where h is height and w is maximum width.
Edge Cases and Tests
Empty tree, one node, skewed tree, duplicate values when reconstruction assumes uniqueness, deep recursion, and tie-breaking for same row/column nodes.
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