Oracle·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Oracle SWE interview that was pretty much all linked list problems. Started simple then escalated fast into K-way merging with complexity analysis. Not the most creative problem set but they clearly wanted to see if you could reason through tradeoffs, not just code.

Questions Asked (2)

Given two sorted linked lists, merge them into a single sorted linked list.

Algorithms & Data Structures

Author's notes

Pretty standard warm-up.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Start by clarifying the problem constraints and edge cases, then propose an iterative two-pointer approach that merges the lists in O(n+m) time and O(1) space. Walk through the algorithm step-by-step, and if time permits, discuss alternative solutions like recursion and their trade-offs.

Pro tip: Demonstrate awareness of production concerns by mentioning that you would handle null inputs gracefully and consider whether modifying the input lists is acceptable, as this shows maturity beyond just solving the algorithm.

1. Clarify requirements and edge cases

Ask about input constraints (e.g., can lists be empty? are they singly or doubly linked? can we modify inputs?) and confirm expected output format. This ensures you understand the problem fully before coding.

2. Outline the two-pointer approach

Explain that you'll use two pointers, one for each list, and a dummy node to build the merged list. Compare the current nodes and append the smaller one to the result, advancing that pointer.

3. Walk through the algorithm with an example

Trace the algorithm on a small example (e.g., 1->3->5 and 2->4->6) to demonstrate correctness and clarify pointer manipulation. This helps catch off-by-one errors early.

4. Analyze complexity and discuss trade-offs

State that the iterative solution runs in O(n+m) time and O(1) space. Mention that a recursive solution is also possible but uses O(n+m) stack space, so iterative is preferred for large lists.

5. Handle edge cases and conclude

Explain how to handle empty lists, lists of different lengths, and duplicate values. Summarize that the dummy node simplifies edge cases and the solution is optimal.

Key Points to Mention

Use a dummy node to simplify edge cases and avoid special handling for the head of the merged list.
Maintain two pointers to traverse the input lists, comparing values and advancing the pointer of the smaller node.
Time complexity is O(n+m) and space complexity is O(1) for the iterative approach.
Recursive solution is elegant but uses O(n+m) stack space, which may cause stack overflow for large lists.
Handle edge cases: one or both lists empty, lists of unequal length, and duplicate values.
Clarify whether the input lists can be modified; if not, create new nodes instead of rearranging pointers.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.

Follow-up: how would you extend the two-list merge to handle K sorted linked lists? Walk through both a divide-and-conquer approach and a min-heap approach, and compare their complexities.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

This is where I stumbled a bit.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Start by restating the problem and clarifying constraints (e.g., K lists, total nodes N). Then describe both approaches: divide-and-conquer by pairwise merging and min-heap by always extracting the smallest head. Finally, compare their time and space complexities, and discuss trade-offs like implementation complexity and constant factors.

Pro tip: Mention that the min-heap approach is often preferred in practice for large K due to its O(N log K) time and O(K) space, but divide-and-conquer can be more cache-friendly and has lower constant factors for small K. Also, note that if the lists are very uneven, a heap can be more efficient.

1. Clarify the problem and constraints

Confirm the number of lists K, total number of nodes N, and whether the lists are sorted. Ask about memory constraints and if in-place merging is required.

2. Explain divide-and-conquer approach

Describe how to merge lists in pairs iteratively (like merge sort) until one list remains. Mention that each merge is O(n) and the total time is O(N log K).

3. Explain min-heap approach

Describe inserting the head of each list into a min-heap, then repeatedly extract the minimum, append to result, and insert the next node from that list. Time is O(N log K), space O(K).

4. Compare complexities and trade-offs

Both have O(N log K) time, but divide-and-conquer uses O(1) extra space (if merging in-place) while heap uses O(K). Discuss constant factors, cache performance, and implementation complexity.

5. Summarize and recommend

Conclude that both are valid; choose based on constraints. For large K, heap may be simpler; for small K or memory-limited, divide-and-conquer is better.

Key Points to Mention

Time complexity: O(N log K) for both approaches, where N is total nodes and K is number of lists.
Space complexity: O(1) for divide-and-conquer (if merging in-place) vs O(K) for min-heap.
Divide-and-conquer merges lists pairwise, reducing the number of lists by half each iteration.
Min-heap approach uses a priority queue to efficiently select the smallest current node among K lists.
Trade-offs: heap has higher constant factors due to heap operations but is easier to implement; divide-and-conquer may be more cache-friendly.
Edge cases: empty lists, K=0, K=1, and lists of vastly different lengths.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.