A block of mass 1.0 kg lies on a rough horizontal floor…

Step 1. One trip. From a stop at $A_1$ to the next stop at $-A_2$ on the other side, friction acts one way only. Energy: $\tfrac12 kA_1^2 - \tfrac12 kA_2^2 = f(A_1 + A_2)$. Divide by $A_1 + A_2$: $A_1 - A_2 = \dfrac{2f}{k} = 4\ \text{cm}$. So each crossing trip shortens the distance by $4\ \text{cm}$.

Step 2. The stops. $15\ \text{cm}$ on the far side, then $11\ \text{cm}$ on the wall side (spring squeezed), then $7\ \text{cm}$ on the far side, then $3\ \text{cm}$ on the wall side. Each time the spring force is more than $2.0\ \text{N}$, so the block sets off again.

Step 3. The last trip. At $3\ \text{cm}$ on the wall side, the spring pushes with $3\ \text{N}$, more than $2.0\ \text{N}$, so the block moves away from the wall. During this trip it swings as if its centre were the point $2\ \text{cm}$ on the wall side, where the spring force just equals friction. It started $3\ \text{cm}$ from the natural position, $1\ \text{cm}$ beyond that centre, so it stops $1\ \text{cm}$ on the other side of the centre: $1\ \text{cm}$ on the wall side. It does not cross the natural length this time. There the spring pushes with only $1\ \text{N}$, less than $2.0\ \text{N}$, so the block stays.

Step 4. Distance and passes. The trips are $26$, $18$, $10$ and $2\ \text{cm}$, in all $56\ \text{cm}$. The first three trips cross the natural length and the last does not, so it passes $3$ times.