My first instinct was just square everything and sort, which is O(n log n) and they immediately asked if I could do better.
Start by clarifying the problem constraints (e.g., input size, sorted order, duplicates) and then propose the two-pointer technique from both ends to achieve O(n) time. Explain that since the array is sorted, the largest squares come from either the leftmost (most negative) or rightmost (most positive) elements, so you can fill the result array from the end. Walk through a small example to demonstrate correctness and edge cases.
Pro tip: Mention that this approach avoids the O(n log n) sort and uses O(n) extra space, which is optimal for the output. Also, relate it to how ML engineers often need to handle sorted data efficiently in preprocessing pipelines.
Ask about input size, whether the array can contain duplicates, and if the output should be a new array or in-place. Confirm that the input is sorted in non-decreasing order.
Recognize that squaring preserves order for non-negative numbers but reverses it for negatives. Use two pointers starting at both ends to pick the larger square and place it at the end of the result array.
Trace the algorithm on a small array like [-4, -2, 0, 1, 3] to show how pointers move and the result is built from the back, ensuring the candidate demonstrates understanding.
State that the time complexity is O(n) because each element is processed once, and space complexity is O(n) for the output array. Mention that this is optimal since we must produce a new array.
Discuss edge cases such as empty array, single element, all negative, all positive, and duplicates. Explain how the algorithm handles them without special cases.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.