Painting With Two Colors

The key is to stop thinking about the two painting operations as equally important. They are not. Blue is painted second, so blue is the boss.

Let the red interval be $A$ with length $a$, and the blue interval be $B$ with length $b$.

After both operations:

• every cell in $B$ is blue;
• every cell in $A$ but not in $B$ is red;
• everything else is white.

So the final blue cells are exactly the interval $B$.

Blue must be centered

In a symmetric coloring, if cell $i$ is blue, then cell $n+1-i$ must also be blue. Since the blue cells form one consecutive interval, that interval must reflect to itself.

A length-$b$ interval can be centered in a row of length $n$ iff the empty space outside it splits evenly between the left and right sides. That means:

$n-b \text{ is even}$

If $n-b$ is odd, blue cannot be symmetric no matter what red does. Red cannot save you here; blue is already final paint. Brutal but fair.

So the first required condition is:

$n-b \equiv 0 \pmod 2$

Now handle red

Assume the blue interval is centered.

There are two cases.

Case 1: $a <= b$

The red interval is no longer than the blue interval. We can simply place the entire red interval inside the blue interval.

Then blue paints over all red cells, so no red remains in the final coloring. The final picture is just:

• white cells on the left,
• one centered blue block,
• white cells on the right.

That is symmetric.

So if $a <= b$, the red part causes zero problems.

Case 2: $a > b$

Now the red interval is longer than the blue interval, so red cannot be fully hidden. Some red cells must survive outside the blue block.

Since the blue block is centered, any surviving red on the left must be matched by surviving red on the right. Because the original red cells came from one consecutive interval, this means the red interval must extend equally past both sides of the centered blue interval.

In other words: the red interval itself must also be centered.

A length-$a$ interval can be centered in $n$ cells iff:

$n-a \text{ is even}$

If this holds, put both intervals centered. Since $a > b$, the red interval contains the blue interval, leaving equal red overhangs on both sides. Symmetric.

If it does not hold, there is no centered red interval of length $a$. Any off-center red interval leaves unmatched red somewhere, which breaks symmetry. No weird hack exists here; the mirror check catches it immediately.

Final condition

We need blue to be centerable, and then red must either be fully hidden or centerable too:

$$(n-b) \text{ is even} \quad \text{and} \quad (a <= b \text{ or } (n-a) \text{ is even})$$

That is the entire problem. One parity check, one size check, done.