0371EasyMath

Combine Two Stock Corrections

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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 stock sheet records every correction to a line as one signed whole number: a positive value adds units to the line, a negative value removes them, and zero leaves the line where it was.

Two corrections have been filed against the same line, first and then second. Whoever filed them wants to replace both with a single correction that moves the line to exactly the same place.

Return that single correction.

Examples

Example 1

Input: first = 7, second = -3
Output: 4

Adding 7 units and then removing 3 leaves the line 4 units above where it started.

Example 2

Input: first = -25, second = -40
Output: -65

Both corrections remove units, so the line ends up 65 units below where it started.

Example 3

Input: first = -100, second = 100
Output: 0

The second correction undoes the first exactly, so the replacement correction is 0.

Example 4

Input: first = 0, second = 58
Output: 58

The first correction leaves the line alone, so only the second one has any effect.

Constraints

-100 <= first, second <= 100

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 sum(first: int, second: int) -> int:

Java

public int sum(int first, int second)

The intended approach

Time O(1)Space O(1)

Corrections are signed shifts of the same line, so applying one and then the other lands wherever the two shifts together land. The replacement correction is therefore the arithmetic total of the two.

Pythonentry point: sum

def sum(first, second):
    # Applying both corrections one after the other shifts the line by their
    # combined amount. O(1) time, O(1) space.
    return first + second

The obvious version

Start from the first correction and step one unit at a time in the direction of the second correction, taking exactly as many steps as its magnitude.

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_sum(first, second):
    total = first
    step = 1 if second > 0 else -1
    for _ in range(abs(second)):
        total += step
    return total

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
4 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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