0108EasyArrayBit Manipulation

The Unpaired Survey Correction

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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.

Two surveyors walk the same stretch of track and file signed elevation corrections in centimetres, negative where the bed sits below the design line. Any correction that truly belongs on the sheet gets filed by both of them, so it shows up on the combined sheet twice. Exactly one correction was filed by a single surveyor, so it shows up there just once.

You are handed the combined sheet as the array corrections, in filing order. Report the value of the correction that shows up only once.

The sheet can run long, so settle it in one sweep over the entries while holding only a fixed amount of extra memory: no tally table, no second copy of the sheet, no reordering.

Examples

Example 1

Input: corrections = [-14, 6, -14, 6, 25]
Output: 25

Both -14 and 6 were filed by each surveyor, leaving 25 as the correction only one of them filed.

Example 2

Input: corrections = [0, -9, -9]
Output: 0

A correction of zero is a real filing, and here it is the one that shows up a single time.

Example 3

Input: corrections = [-30000]
Output: -30000

A sheet with a single entry can only be the lone filing, so it is reported as is.

Constraints

1 <= corrections.length <= 3 * 10^4
-3 * 10^4 <= corrections[i] <= 3 * 10^4
Every value on the sheet shows up exactly twice apart from one value, which shows up exactly once.

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 unpaired_correction(corrections: list[int]) -> int:

Java

public int unpairedCorrection(int[] corrections)

The intended approach

Time O(n)Space O(1)

Fold the whole sheet together with bitwise exclusive-or. Any value filed twice cancels itself out bit by bit regardless of where the two filings sit, so what is left after the sweep is exactly the value filed once. This holds for negatives too, since the sign bit cancels the same way as any other bit, and it needs a single accumulator.

Pythonentry point: unpaired_correction

def unpaired_correction(corrections):
    residue = 0
    for value in corrections:
        residue ^= value
    return residue

The obvious version

For every entry, count how many times its value shows up anywhere on the sheet with a second loop, and report the first entry whose count is one. O(n^2).

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_unpaired_correction(corrections):
    for value in corrections:
        seen = 0
        for other in corrections:
            if other == value:
                seen += 1
        if seen == 1:
            return value
    return 0

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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