0689MediumArrayMathTwo PointersSorting

Sorted Dial Responses

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A tuning rig is swept through a list of dial offsets, given in non-decreasing order and possibly containing repeats. At offset x the rig reports the response

a * x * x + b * x + c

Return the responses of all the offsets, arranged in non-decreasing order. The returned list has one entry per offset, so a repeated offset contributes its response more than once. Any of a, b and c may be zero or negative.

Examples

Example 1

Input: offsets = [-1, 0, 1, 2, 3], a = 1, b = -4, c = 3
Output: [-1, 0, 0, 3, 8]

The five responses are 8, 3, 0, -1 and 0 in dial order, and the returned list is those same five numbers arranged from smallest to largest.

Example 2

Input: offsets = [-4, -2, 0, 1], a = 0, b = -3, c = 1
Output: [-2, 1, 7, 13]

With a zero leading coefficient the responses are 13, 7, 1 and -2, so the returned list reverses the dial order.

Example 3

Input: offsets = [-2, -2, 3, 3], a = 1, b = 1, c = -6
Output: [-4, -4, 6, 6]

Offset -2 reports -4 and offset 3 reports 6. Each offset appears twice, so each response appears twice in the answer.

Constraints

1 <= offsets.length <= 200
-100 <= offsets[i], a, b, c <= 100
offsets is given in non-decreasing order.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def sorted_responses(offsets: list[int], a: int, b: int, c: int) -> list[int]:

Java

public int[] sortedResponses(int[] offsets, int a, int b, int c)

The intended approach

Time O(n) for n offsets: each cursor moves once per offset, no sorting.Space O(n) for the returned list, O(1) besides it.

The response is a quadratic in the offset, so along a non-decreasing list of offsets the extreme response always sits at one of the two ends of the remaining stretch. Keep a cursor at each end and repeatedly take the more extreme of the two, writing the answer from its back when the leading coefficient is positive and from its front when it is negative. A zero leading coefficient leaves a straight line, whose extremes are also at the ends, so it can travel the same path as the positive case.

Pythonentry point: sorted_responses

def sorted_responses(offsets, a, b, c):
    n = len(offsets)
    out = [0] * n                      # O(n) for the returned list

    def response(x):
        return a * x * x + b * x + c

    lo = 0
    hi = n - 1
    if a >= 0:
        # with a positive leading term the two ends of the offset list carry the
        # largest responses, so fill the answer from its back; with a zero
        # leading term the response is a straight line and its extremes still sit
        # at the two ends, so the same sweep works
        for pos in range(n - 1, -1, -1):
            left = response(offsets[lo])
            right = response(offsets[hi])
            if left >= right:
                out[pos] = left
                lo += 1
            else:
                out[pos] = right
                hi -= 1
    else:
        # with a negative leading term the ends carry the smallest responses
        for pos in range(n):
            left = response(offsets[lo])
            right = response(offsets[hi])
            if left <= right:
                out[pos] = left
                lo += 1
            else:
                out[pos] = right
                hi -= 1
    return out

The obvious version

Evaluate the response at every offset into a list and hand that list to a general purpose sort.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_sorted_responses(offsets, a, b, c):
    values = []
    for x in offsets:
        values.append(a * x * x + b * x + c)
    values.sort()
    return values

How this problem was checked

Reference solution executed against 13 test cases — 13 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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