Use the two-pointer technique: advance a fast pointer k nodes ahead, then move both pointers until the fast pointer reaches the end. The slow pointer will then be at the k-th node from the end. This achieves O(n) time and O(1) space in a single pass.
Pro tip: Clarify edge cases upfront (e.g., k > list length, k = 0, empty list) and discuss how to handle them, showing attention to detail. Also, mention that if the list length is known, a two-pass approach is simpler, but the two-pointer method is optimal for unknown length.
Restate the problem in your own words and ask clarifying questions about input constraints, edge cases, and expected return value (e.g., return the node or its value).
Compare naive two-pass (compute length, then traverse) vs. optimal two-pointer (single pass). Explain trade-offs in time and space complexity.
Initialize two pointers at head. Move the first pointer k steps forward. Then move both pointers one step at a time until the first pointer reaches null. The second pointer is the answer.
Consider cases: empty list, k <= 0, k > length. Decide on behavior (e.g., return null or throw exception) and incorporate checks.
State time O(n) and space O(1). Walk through a small example to verify correctness, and mention potential follow-ups like modifying the list or handling circular lists.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Clarify the problem constraints (e.g., input size, value range, tie-breaking) and then propose a solution using a hash map to count frequencies, followed by a heap or bucket sort to extract the top k elements. Discuss trade-offs between different approaches to demonstrate depth.
Pro tip: Mention that you would first ask about constraints (e.g., n and k sizes, whether k is always valid) to choose the optimal algorithm, and explicitly state the time and space complexity of your chosen approach.
Ask about input size, value range, whether k is always valid, and how to handle ties. This ensures you choose the right algorithm and avoid edge-case bugs.
Use a hash map to count the occurrence of each element. This takes O(n) time and O(n) space.
Use a min-heap of size k to efficiently find the k most frequent elements, or use bucket sort if the frequency range is small. Discuss the trade-offs.
State the time and space complexity of your solution. Consider edge cases like k=0, k equal to the number of unique elements, or all elements having the same frequency.
Walk through a small example to verify correctness, and mention potential optimizations or alternative approaches.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Same heap pattern as the previous question basically, just with Euclidean distance as the comparator.
Start by clarifying the problem: confirm whether points are unique, if k is valid, and if distance is Euclidean. Then propose a max-heap of size k to efficiently find the k closest points in O(n log k) time, or a quickselect approach for O(n) average time. Discuss trade-offs and handle edge cases like k >= n.
Pro tip: Mention that you can avoid computing square roots by comparing squared distances, which is a common optimization and shows attention to performance. Also, be prepared to discuss how you would handle duplicate points or ties in distance.
Ask about input size, whether k is always valid, if points can be duplicated, and if distance is Euclidean. Confirm expected output format (list of points or indices).
Decide between sorting (O(n log n)), max-heap (O(n log k)), or quickselect (O(n) average). Explain why max-heap is often optimal for large n and small k.
Write code for the chosen approach. For max-heap, iterate through points, push negative squared distance and point, and pop when heap size exceeds k. For quickselect, partition based on distance.
State time and space complexity. Discuss edge cases: k=0, k>=n, empty list, duplicate points, and points with same distance.
Walk through a small example to verify correctness. Mention potential optimizations like early termination or using a balanced tree if needed.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use a stack to process characters one by one: if the stack top matches the current character, pop it; otherwise, push the character. After processing all characters, the stack contains the final string with all adjacent duplicates removed.
Pro tip: Clarify whether collisions can cascade (e.g., after removal, new adjacent duplicates form) and confirm that the stack approach naturally handles cascading because it processes left to right and always compares with the current top. Also, mention that the stack can be implemented as a string or list for efficiency.
Restate the problem to ensure clarity: adjacent identical characters annihilate each other, and this process repeats until no more collisions occur. Confirm with examples like 'abbba' -> 'aa' -> ''.
Select a stack (or a string builder) to efficiently track characters and handle collisions in O(n) time. Explain why a stack is ideal: it allows O(1) push/pop and automatically handles cascading removals.
Iterate through each character: if the stack is not empty and the top equals the current character, pop; else push. After the loop, the stack contains the final string.
State that time complexity is O(n) because each character is pushed and popped at most once, and space complexity is O(n) for the stack in the worst case.
Walk through examples: empty string, no collisions, all same characters, and cascading collisions. Verify the algorithm handles them correctly.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.