0069EasyArrayTwo PointersSorting

Sorted Drift Energies

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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 calibration rig logs how far each sensor sits from its target. The log is drift, where a negative entry means the sensor reads low, a positive entry means it reads high, and the entries are already arranged from lowest to highest, so ties may sit next to each other.

The rig grades a sensor by its drift energy, the entry multiplied by itself, which wipes out the sign. Produce the drift energy of every entry, arranged from smallest to largest.

Sorting the energies afterwards works but throws away what the log already tells you. Aim for one pass over drift.

Examples

Example 1

Input: drift = [-9, -4, 0, 2, 7]
Output: [0, 4, 16, 49, 81]

The energies are 81, 16, 0, 4 and 49; arranged from smallest to largest they read 0, 4, 16, 49, 81. The largest energy came from the leftmost entry.

Example 2

Input: drift = [-8, -5, -3, -1]
Output: [1, 9, 25, 64]

Every sensor reads low, so squaring flips the arrangement end for end.

Example 3

Input: drift = [-2, -2, 0, 2, 2]
Output: [0, 4, 4, 4, 4]

Four entries share the energy 4, and each of them is reported.

Constraints

1 <= drift.length <= 10^4
-10^4 <= drift[i] <= 10^4
drift is arranged from lowest to highest value

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 drift_energies(drift: list[int]) -> list[int]:

Java

public int[] driftEnergies(int[] drift)

The intended approach

Time O(n)Space O(n)

Because the log runs from lowest to highest, the largest energy always belongs to one of the two end entries: the most negative one or the most positive one. Keep a cursor at each end and fill the answer from its last slot backwards, each step taking whichever end has the bigger energy and stepping that cursor inward. Every entry is placed exactly once, so the answer comes out in one pass with no sorting.

Pythonentry point: drift_energies

def drift_energies(drift):
    size = len(drift)
    energies = [0] * size
    left = 0
    right = size - 1
    for slot in range(size - 1, -1, -1):
        low = drift[left] * drift[left]
        high = drift[right] * drift[right]
        if low > high:
            energies[slot] = low
            left += 1
        else:
            energies[slot] = high
            right -= 1
    return energies

The obvious version

Square every entry into a fresh list and hand it to a general sort, ignoring the arrangement the log already had.

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_drift_energies(drift):
    squared = []
    for value in drift:
        squared.append(value * value)
    squared.sort()
    return squared

How this problem was checked

Reference solution executed against 10 test cases — 10 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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