The naive approach is obvious, square everything and sort it, but that's O(n log n) and they pushed for better.
Start by clarifying the problem and discussing a naive approach that squares each element and sorts, which takes O(n log n) time. Then, leverage the fact that the input is sorted to design an O(n) two-pointer solution that builds the result from largest to smallest. Walk through an example to demonstrate correctness and analyze time/space complexity.
Pro tip: Emphasize that the two-pointer approach works because the largest squared values must come from either end of the sorted array, and building the result from the end avoids extra reversals. Mention edge cases like negative numbers and duplicates to show thoroughness.
Restate the problem, ask about input constraints (e.g., can elements be negative? duplicates? empty array?), and confirm expected output format.
Describe the straightforward solution: square each element and sort the result. Analyze its time complexity as O(n log n) and note it doesn't use the sorted property.
Explain that since the array is sorted, the largest squared values are at the ends. Use two pointers starting at both ends, compare absolute values, and place the larger square at the end of the result array, moving pointers inward.
Trace the algorithm on a sample array (e.g., [-4, -1, 0, 3, 10]) to show how the result is built in O(n) time and O(n) space.
State time complexity O(n) and space O(n) for the output. Discuss edge cases: empty array, all negatives, all positives, duplicates, and large values (overflow considerations).
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