My first instinct was to just map the formula and re-sort at the end, which works but they clearly wanted something smarter.
First, clarify the problem: the quadratic formula is f(x) = ax^2 + bx + c, and we need to apply it to each element of a sorted array, then return the results sorted. Recognize that if a > 0, the function is convex (U-shaped), so the sorted order of outputs can be obtained by merging two sorted sequences: one from the left end (decreasing) and one from the right end (increasing). If a < 0, the function is concave (inverted U), so the maximum is at the vertex, and the sorted order can be obtained by splitting at the vertex and merging. If a = 0, it's linear, so order is preserved or reversed depending on b. Implement an O(n) two-pointer approach to merge the two sorted halves.
Pro tip: Mention that this is a common pattern for applying monotonic functions to sorted arrays, and that handling the vertex (where the function changes direction) is key to achieving O(n) time. Also, note that if the array contains negative numbers, the vertex may lie within the array, so you need to find the split point where the function's derivative changes sign.
Ask whether the quadratic formula means f(x) = ax^2 + bx + c with given coefficients, and confirm that the input array is sorted in ascending order. Also, check if the array can contain negative numbers and if the coefficients are fixed.
Determine the shape of the quadratic: if a > 0, it's convex with a minimum at the vertex; if a < 0, it's concave with a maximum at the vertex. The vertex x = -b/(2a) divides the array into two monotonic parts.
Use binary search to find the index where the array values cross the vertex, or simply find the point where the function's derivative changes sign. This splits the array into two parts: one where f is decreasing and one where f is increasing (or vice versa).
Apply f to each element in both parts. The left part (if decreasing) will produce values in decreasing order, and the right part (if increasing) will produce values in increasing order. Use two pointers to merge these into a single sorted array in O(n) time.
Consider cases where a = 0 (linear function), the vertex is outside the array, or the array has duplicates. Discuss the time complexity (O(n) after O(log n) for finding the vertex) and space complexity (O(n) for the output).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.