Watermelon

The whole problem comes down to parity plus one annoying edge case.

Suppose the two part weights are $a$ and $b$. We need

$a+b=w,$

where both $a$ and $b$ are positive even integers.

Necessary conditions

The sum of two even integers is even. Therefore, if $w$ is odd, no valid split can exist.

That does not mean every even $w$ works. The smallest positive even integer is $2$, so both boys receiving a positive even part requires

$w \ge 2+2=4.$

This rules out $w=2$. That is the only trap here: splitting it as $0+2$ has even weights, but zero is not positive, so Pete or Billy gets existential disappointment instead of watermelon.

Sufficient condition

Now suppose $w$ is even and $w>2$. Since integer weights are used, this means $w\ge4$. Choose the parts

$2 \quad\text{and}\quad w-2.$

The first part is positive and even. Because $w$ and $2$ are even, $w-2$ is also even; and because $w\ge4$, it is positive. Thus this split is always valid.

So the exact answer is:

• print YES when $w$ is even and greater than $2$;
• otherwise print NO.

Correctness proof

If the algorithm prints YES, then $w>2$ and $w$ is even. The split $(2,w-2)$ consists of two positive even integers, so a valid division exists.

If a valid division exists, both parts are positive even integers. Their sum $w$ must therefore be even, and each part is at least $2$, giving $w\ge4$, hence $w>2$. Therefore the algorithm prints YES.

The algorithm prints YES exactly when a valid division exists.

Complexity

The algorithm uses $O(1)$ time and $O(1)$ extra space.