Merge Two Sorted Linked Lists
Merge two sorted linked lists and return it as a new sorted list. The new list should be made by splicing together the nodes of the first two lists.
Why Interviewers Ask This
Stripe interviewers ask this to evaluate a candidate's mastery of pointer manipulation and edge case handling in low-level data structures. They specifically look for the ability to maintain list integrity while traversing without creating unnecessary memory overhead. This problem tests logical precision, as incorrect pointer updates can easily cause infinite loops or data loss, reflecting the reliability required for financial transaction systems.
How to Answer This Question
1. Clarify Requirements: Confirm if you can modify existing lists or must create a new one, and handle null inputs immediately.
2. Visualize the Process: Mentally trace how nodes from both lists interleave, identifying the smallest head node first.
3. Choose a Strategy: Opt for an iterative dummy-head approach to simplify boundary conditions rather than recursion, which avoids stack overflow risks on large inputs.
4. Implement Logic: Initialize a dummy node and a current pointer. Loop while both lists have nodes, comparing values to splice the smaller node into your result chain.
5. Handle Remainders: After the loop, attach any remaining non-empty list directly to the end, as it is already sorted.
6. Validate Edge Cases: Explicitly test scenarios where one list is empty, both are empty, or all elements are identical to ensure robustness expected at Stripe.
Key Points to Cover
- Using a dummy head node to eliminate complex boundary condition checks
- Maintaining O(n + m) time complexity and O(1) space complexity
- Correctly handling cases where one list becomes null before the other
- Demonstrating precise pointer manipulation without creating new nodes unnecessarily
- Explicitly stating assumptions about input validity and modification constraints
Sample Answer
To merge two sorted linked lists efficiently, I would use an iterative approach with a dummy head node to streamline the logic. First, I'd initialize a dummy node pointing to null and a current pointer set to this dummy. This dummy node acts as a placeholder, eliminating the need for special cases when attaching the very first node.
Next, I would iterate through both lists simultaneously. In each step, I compare the values of the current nodes in list A and list B. If list A's value is smaller, I link the current pointer to list A's node and advance list A's pointer; otherwise, I do the same for list B. After linking, I move the current pointer forward. This ensures the merged list remains sorted by splicing nodes directly from the input lists.
Once the loop terminates because one list is exhausted, I simply attach the remaining nodes of the non-empty list to the end of my merged list, since that remainder is guaranteed to be larger than everything already processed. Finally, I return the next node of the dummy head, which is the true start of the merged list. This approach runs in O(n + m) time with O(1) space complexity, ensuring high performance and memory efficiency, which aligns well with Stripe's focus on scalable infrastructure.
Common Mistakes to Avoid
- Failing to handle null pointers, causing runtime errors when one list is shorter
- Creating new nodes instead of splicing existing ones, violating the problem constraints
- Using recursion without considering stack depth limits for very long lists
- Losing track of the head pointer due to missing a dummy node wrapper
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