Can You Split a Watermelon Evenly? | Codeforces 4A Explained

A quick Python solution to one of the most iconic beginner problems on Codeforces

Here’s a simple but classic problem from Codeforces that’s perfect for beginners —
it tests your understanding of even numbers and edge conditions in just a few lines of code.

The challenge? Decide if a given number can be split into two positive even integers.

Try it yourself here:
Codeforces 4A – Watermelon

Problem breakdown

The goal is simple:

Given a watermelon of weight w, can we split it into two parts such that both parts are positive even integers?

Let’s clarify the requirements:

• Both parts must be even numbers — no odd-number leftovers allowed.

• Both must be greater than 0 — no empty slices.

• They don’t have to be equal in weight.

For example:

• 8→YES(2+6,4+4, etc.)

• 6→YES(2+4)

• 3→NO(Can’t split into two even numbers)

The trick here is realizing that:

Any even number greater than 2 can be split into two positive even numbers.

So really, the only invalid even number is 2 — too small to divide into two positive even integers.

My python solution

def dividable(w):
    for i in range(2,w,2):
        if i%2==0 and (w-i)%2==0:
            return "YES"
        else:
            continue
    return "No"

if __name__ == "__main__":
    w = int(input(""))
    print(dividable(w))

Here’s the first version of my solution:

This code loops through all even numbers from 2 up to w - 2, and checks whether both i and w - i are even.
If it finds such a pair, it immediately returns "YES".
If the loop ends without finding any valid combination, it returns "No".

It works — but after understanding the underlying logic, I realized it can actually be simplified quite a lot.

Cleaner Version

Once you know that any even number greater than 2 is splittable, the logic becomes crystal clear:

def dividable(w):
    return "YES" if w > 2 and w % 2 == 0 else "No"

This one-liner checks two conditions:

1. The weight w must be even (w % 2 == 0)

2. It must be greater than 2 (because 2 is too small to split into two positive even numbers)

Time and Space Complexity

• Time Complexity

  • First version: O(n) — due to the loop

  • Second version: O(1) — constant time check

• Space Complexity

  • Both versions use O(1) space

Ham Tori’s Takeaway

Don’t let simple problems fool you — they’re often a great way to sharpen your thinking.
This one may seem easy at first glance, but it teaches an important skill:
how to spot patterns in constraints and reduce logic to its cleanest form.

See you in the next problem!