Classic two-pointer problem once you see it, but I fumbled around for a bit thinking about just squaring everything and sorting after.
Start by clarifying the problem constraints and edge cases, then propose a two-pointer approach that leverages the sorted order to build the result in O(n) time. Explain why squaring and sorting naively is O(n log n) and how the two-pointer method improves it. Walk through a small example to demonstrate correctness.
Pro tip: Mention that the largest squares come from either end of the array, so comparing absolute values from both ends lets you fill the result from the back in one pass. This shows you understand the structure of the problem, not just the algorithm.
Ask about input size, whether the array can contain negative numbers, and if duplicates or empty arrays are possible. Confirm that the output should be a new array, not in-place.
Mention the straightforward solution: square each element and sort the result, which takes O(n log n) time. Acknowledge it works but isn't optimal given the sorted input.
Explain that since the array is sorted, the largest squared values are at the extremes. Use two pointers starting at both ends, compare absolute values, and place the larger square at the end of the result array, moving inward.
Trace the algorithm on a sample input like [-4, -1, 0, 3, 10] to show how the result is built from the back, ensuring the candidate demonstrates understanding.
State that time complexity is O(n) and space complexity is O(n) for the output array. Discuss edge cases like empty array, all negatives, all positives, and zeros.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.