I went straight for the min-heap approach and it landed fine.
Start by clarifying the problem constraints and edge cases, then discuss multiple approaches such as sequential merging, divide-and-conquer, and heap-based merging. Compare their time and space complexities, and implement the most efficient one (heap-based or divide-and-conquer) with clean code and thorough testing.
Pro tip: Mention that a heap-based approach is optimal for large k, but divide-and-conquer is often preferred in practice for its simplicity and lower constant factors. Also, discuss how to handle edge cases like empty lists and duplicate values.
Ask about the number of lists (k), average length, whether lists can be empty, and if the merged list should be a new list or modify existing nodes. Confirm the expected time and space complexity.
Outline at least two approaches: (1) sequentially merging lists one by one, (2) using a min-heap to efficiently select the smallest node among all lists, and (3) divide-and-conquer by merging pairs of lists. Compare their complexities.
Select the most efficient approach based on constraints. For example, if k is large, heap-based O(N log k) is optimal; if k is small, divide-and-conquer may be simpler. Explain your reasoning.
Write clean code for the chosen approach. Use a dummy node to simplify list construction, and handle edge cases like empty input. For heap-based, store (value, list_index, node) tuples.
Walk through test cases: empty lists, single list, lists with different lengths, and duplicate values. Analyze time and space complexity, and discuss potential optimizations or trade-offs.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This tripped me up more than it should have.
Use BFS to traverse the tree level by level, tracking each node's column index. Store nodes in a map keyed by column, and for each column, maintain the order of insertion to preserve top-to-bottom and left-to-right ordering. Finally, output the columns from leftmost to rightmost.
Pro tip: Clarify the ordering rule: within the same row and column, nodes should appear left to right. This means when processing a level, enqueue left child before right child, and process nodes in the order they are dequeued. Also, mention that using a TreeMap or sorting the keys at the end ensures columns are ordered correctly.
Restate the problem to ensure understanding: return nodes column by column, top to bottom, and left to right for ties. Discuss edge cases like empty tree, single node, and skewed trees.
Explain that BFS naturally processes nodes level by level, ensuring top-to-bottom order. Use a queue of (node, column) pairs and a map from column to list of node values.
While the queue is not empty, dequeue a node, append its value to the list for its column, and enqueue its left child with column-1 and right child with column+1. This preserves left-to-right order for same row and column.
After BFS, extract the lists from the map in sorted order of columns (e.g., using a TreeMap or sorting keys). Return the combined list of values.
State time and space complexity: O(N) time and O(N) space. Walk through a small example to verify correctness, and mention potential optimizations if needed.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.