Merge K Sorted Lists
Last updated: February 21, 2025
Quick Overview
Merge k sorted linked lists into one sorted linked list and return the head of the merged list.
Rivian
February 21, 20258
7
1,847 solved
Merge k sorted linked lists into one sorted linked list and return the head of the merged list.
This classic heap problem appears in Rivian coding rounds. It models real scenarios like merging sorted telemetry streams from multiple vehicle sensors. The interviewer evaluates your heap usage, code cleanliness, and complexity analysis.
What the Interviewer Expects
- Use a min-heap to efficiently track the smallest element across all lists
- Handle edge cases including empty lists and lists of different lengths
- Achieve optimal O(N log k) time complexity where N is total elements
- Write clean code with a dummy head node to simplify list building
- Compare with the divide-and-conquer alternative approach
Key Topics to Cover
How to Approach This
- Clarify input constraints and edge cases before writing code.
- Walk through your approach verbally and confirm with the interviewer before coding.
- Start with a brute force solution, then optimize. Mention time and space complexity.
- Test your solution with examples, including edge cases like empty input or duplicates.
- Consider common patterns: sliding window, two pointers, hash map, BFS/DFS, dynamic programming.
Possible Follow-up Questions
- What is the time and space complexity of your solution?
- How would you solve this if the input were sorted arrays instead of linked lists?
- Can you solve this using divide and conquer? What are the tradeoffs?
- How would you handle this in a streaming context where lists grow over time?
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Practice DSA ProblemsSample Answer
Heap Approach
Initialize a min-heap with the head node of each non-empty list. Use a dummy head node to simplify result list construction. Repeatedly extract the mi...
Implementation Details
Create a dummy node and a current pointer. Push (node.val, index, node) tuples to the heap, where index breaks ties for nodes with equal values (since...
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