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Oracle·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Apr 2026

Summary

Oracle SWE interview with a linked list merging problem that had a twist I wasn't fully expecting. The core algorithm wasn't too bad but the tie-breaking rule added a layer that tripped me up at first.

Questions Asked (1)

Q1

Given k sorted singly linked lists where each node holds a key-value pair, merge them into a single sorted list. If the same key appears in multiple lists, the value from the higher-indexed list should win. What's your approach and complexity?

Algorithms & Data Structures
Author's notes

My first instinct was just a standard k-way merge with a min-heap and I started coding that up before fully reading the tie-breaking part.

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AI HintsAI Generated

Suggested Approach

Use a min-heap (priority queue) to efficiently merge the k sorted lists by always extracting the smallest key. For duplicate keys, prioritize the node from the higher-indexed list by comparing list indices when keys are equal. Then analyze time and space complexity, noting the heap size is at most k.

Pro tip: Mention that if the lists are very large and k is small, a heap is optimal, but if k is huge, a divide-and-conquer merge might be better; also clarify how you handle duplicate keys to ensure the higher-indexed list wins.

1. Clarify requirements and edge cases

Confirm that lists are sorted by key, keys may be duplicated, and higher-indexed list wins on ties. Discuss edge cases like empty lists, k=0, or all lists empty.

2. Choose data structure

Select a min-heap (priority queue) to store the current head of each list. The heap comparator should order by key, and for equal keys, by list index descending (so higher index is extracted first).

3. Initialize and merge

Insert the head of each non-empty list into the heap. Repeatedly extract the minimum node, append it to the result list, and if that node has a next, insert the next node into the heap.

4. Handle duplicates correctly

When extracting, if multiple nodes have the same key, the heap comparator ensures the one from the higher-indexed list is extracted first. This automatically enforces the 'higher-indexed list wins' rule.

5. Analyze complexity

Time complexity: O(N log k) where N is total number of nodes and k is number of lists. Space complexity: O(k) for the heap (plus O(N) for the output list).

Key Points to Mention

  • Min-heap (priority queue) with custom comparator to handle duplicate keys and list indices.
  • Time complexity O(N log k) and space complexity O(k) for the heap.
  • Handling of duplicate keys: higher-indexed list wins, achieved by comparator ordering.
  • Edge cases: empty lists, k=0, all lists empty, and lists of varying lengths.
  • Alternative approaches: divide-and-conquer merge (O(N log k) time, O(1) extra space if done iteratively) and their trade-offs.
  • Stability and correctness: ensuring the merged list is sorted and duplicates are resolved as specified.

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