Pathless

Let

$S=\sum_{i=1}^{n} a_i.$

This is the sum Alice gets from the straight path

$1\to2\to3\to\cdots\to n.$

On a line, any walk from $1$ to $n$ must visit every index at least once. Since every $a_i\ge 0$, Alice can never get a sum smaller than $S$.

So if $s<S$, Bob can output any rearrangement. Alice cannot go below the mandatory sum.

If $s=S$, Bob cannot stop Alice, because the straight path already works.

What extra movement does

Suppose Alice is at index $i$ and wastes time by going

$i\to i+1\to i.$

That detour adds

$a_{i+1}+a_i.$

Similarly, going left and then right across the same edge adds the same value. Therefore, after the mandatory straight path, every extra contribution is a nonnegative combination of adjacent pair sums:

$a_1+a_2,\ a_2+a_3,\ \ldots,\ a_{n-1}+a_n.$

Conversely, Alice can realize any such combination by doing the needed back-and-forth detours before continuing to the right. So this characterization is exact, not just hand-wavy vibes.

Now values are only $0,1,2$, so adjacent pair sums can be:

$0,1,2,3,4.$

The important one is $1$. It happens only when a $0$ is adjacent to a $1$.

If some adjacent pair has sum $1$, Alice can add $1$ as many times as she wants. Then every target $s\ge S$ is reachable. Bob definitely does not want that.

So to block $S+1$, Bob should avoid every adjacent $0,1$ pair. A clean arrangement is:

$0,0,\ldots,0,\ 2,2,\ldots,2,\ 1,1,\ldots,1.$

Its adjacent pair sums are only:

• $0+0=0$,
• $0+2=2$,
• $2+2=4$,
• $2+1=3$,
• $1+1=2$.

There is no way to add exactly $1$. So if $s=S+1$, this arrangement works for Bob.

Why larger targets cannot be blocked

Take any arrangement.

If it has a neighboring $0$ and $1$, Alice can add $1$, so all sums $S,S+1,S+2,\ldots$ are reachable.

Otherwise, there is no adjacent $0,1$ pair. Since the array contains at least one $0$, one $1$, and one $2$, any path in the array from a $0$ position to a $1$ position must pass through $2$s. That means somewhere there is a $0$ next to a $2$, and somewhere there is a $2$ next to a $1$.

So Alice can add both:

$0+2=2$

and

$2+1=3.$

Using $2$ and $3$, Alice can make every integer at least $2$:

• even $d$: use $d/2$ copies of $2$,
• odd $d\ge 3$: use one $3$, then fill the rest with $2$s.

So every target $S+d$ with $d\ge 2$ is always reachable. Bob cannot block those.

Final cases:

• If $s<S$: output any rearrangement.
• If $s=S$: output $-1$.
• If $s=S+1$: output all $0$s, then all $2$s, then all $1$s.
• If $s>S+1$: output $-1$.

That is the whole trick. No DP needed; DP would be using a sledgehammer on a gummy bear.