Rabbits

The trick is to stop thinking about rabbits as cute little chaos machines and start thinking about what can actually make one move.

A rabbit at an empty pot chooses left or right. It does not jump if:

• it looks outside the array,
• it looks at another empty pot, because that pot has a rabbit,
• or it looks at a flower, but another rabbit from the opposite side also looks at that same flower.

So the only dangerous move is:

a rabbit points into a flower, and nobody points into that flower from the other side.

Everything else is automatically safe.

Turn directions into variables

Only positions with s[i] = 0 contain rabbits. For each such position, define a variable:

• 0 means the rabbit looks left,
• 1 means the rabbit looks right.

Now inspect what constraints exist.

Suppose we have this pattern:

0 1 0

There are two rabbits around one flower.

If the left rabbit points right into the flower, then the right rabbit must point left into the same flower. Otherwise the left rabbit jumps.

If the right rabbit points left into the flower, then the left rabbit must point right too. Otherwise the right rabbit jumps.

So around every 010, the valid direction pairs are:

L R   both point away from the flower
R L   both point into the flower

The invalid pairs are:

L L   only right rabbit attacks the flower
R R   only left rabbit attacks the flower

That means the two direction variables must be different.

So every substring 010 gives an inequality constraint:

$x_i \ne x_{i+2}$

where both endpoints are zero positions.

Forced directions

Now consider a rabbit next to a flower with no rabbit on the other side:

0 1 1

The left rabbit cannot point right, because it would target the flower and there is no rabbit on the opposite side to block it. Therefore it is forced to point left.

Similarly:

1 1 0

The right rabbit cannot point left, so it is forced to point right.

This also covers borders. In 01 at the end, if the flower has no zero on its other side, the zero cannot point into it. Pointing outward or toward another rabbit is fine.

For every zero position i:

• If i - 1 is a flower and i - 2 is not a zero position, then x_i cannot be left, so force x_i = right.
• If i + 1 is a flower and i + 2 is not a zero position, then x_i cannot be right, so force x_i = left.

If one rabbit is forced both left and right, answer is immediately NO.

Now it is just graph coloring

Make a graph whose vertices are the zero positions.

For every pattern 010, connect the two zero positions with an edge meaning:

$x_u \ne x_v$

Some vertices may already have a forced color.

Because edges only connect positions distance $2$ apart, the graph is basically a bunch of chains. No cursed 2-SAT monster needed. We can still write it as a normal BFS/DFS coloring check:

1. Start from every unvisited zero position.
2. If it has a forced value, use that. Otherwise choose either value, say 0.
3. Across every edge, assign the opposite value.
4. If we ever assign a vertex a value different from what it already had, answer NO.
5. If a forced value disagrees with the BFS value, answer NO.

If all components can be colored, answer YES.

Why this is sufficient

We handled every possible way a rabbit might move:

• Target is outside: safe.
• Target is a zero: safe because there is a rabbit there.
• Target is a one:
  • if there is no zero on the other side, we forbid pointing there,
  • if there is a zero on the other side, the 010 inequality guarantees either both rabbits point into the flower or neither does.

So no rabbit can make an unblocked jump. That is exactly what the problem asks.

Complexity is linear: $O(n)$ per test case, and $O(\sum n)$ overall. Memory is also $O(n)$.