1-1

Observation

Let $l$ be the first index with $s_l=\texttt{1}$, and let $r$ be the last one.

If there is no $\texttt{1}$, no move can ever happen, so the answer is simply:

$$0\ 0$$

Now assume at least one $\texttt{1}$ exists.

Positions outside $[l,r]$ can never become $1$. A position can only be changed if it has $1$'s on both sides:

$$s_{i-1}=s_{i+1}=\texttt{1}.$$

Outside the first/last $1$, one side has no available $1$ at all. So all action is trapped inside $[l,r]$.

Maximum

Inside $[l,r]$, we can fill everything with $1$'s. Whenever two $1$'s are separated, the operation can gradually create more $1$'s between them. So eventually the whole segment becomes all $1$'s.

Thus:

$$\text{max}=r-l+1.$$

Minimum

For the minimum, we want as few surviving $1$'s as possible, but $s_l$ and $s_r$ cannot be changed if they are endpoints of the whole reachable segment.

Between surviving $1$'s, the gap cannot exceed $2$. If two remaining $1$ positions were distance $3$ or more apart, the middle area could not be supported by moves correctly. The sparsest valid pattern is basically:

$$\texttt{101010}\cdots$$

So if $d=r-l$, we need the fewest jumps of length at most $2$ to cover distance $d$:

$$\left\lceil \frac d2 \right\rceil$$

jumps, meaning:

$$\text{min}=\left\lceil \frac{r-l}{2} \right\rceil+1.$$

In integer C++ arithmetic:

$$\left\lceil \frac d2 \right\rceil = \frac{d+1}{2}.$$

Complexity

We scan the string once.

$$O(n)$$

per test case, with $O(1)$ extra memory. Constraints are tiny, but the solution is still clean.