Instead of choosing every carpet's position, consider the carpet laid first: it lies below every other carpet.

Suppose this is carpet $k$, and its visible section is $[x,y)$. No other carpet can intersect this section. Any such carpet would lie above carpet $k$ and hide it there.

Each other carpet is a single interval, so it lies entirely to the left or entirely to the right of $[x,y)$. Its required visible position determines which side:

• Carpets with smaller labels lie entirely to the left.
• Carpets with larger labels lie entirely to the right.

Moreover, each side is a valid display by itself. Every point on that side already shows one of its own carpets, so removing carpet $k$ does not uncover a gap.

This suggests an interval state. Let $dp[l][r]$ be the minimum corridor length for displaying carpets $l,l+1,\ldots,r$ in that order. Define $dp[l][r]=0$ when $l>r$.

If carpet $k$ is the lowest carpet of this subproblem, write $L=dp[l][k-1]$ and $R=dp[k+1][r]$. Any such display has length at least

$$\max(a_k,\;L+R+1).$$

The whole carpet $k$ must fit, giving the first bound. Its two sides need at least $L$ and $R$ units, and its own visible section needs at least one unit, giving the second.

The remaining question is whether these two bounds are enough.