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In this case, one realises that the horizontal length constraint implies that for a given $N$, $r$ is constrained as:
$$
r=\frac{L-3p-Np}{4+2N}
$$
Then for this given $N$ and $r$, one may calculate $\ell_{resid}$ and subsequently $s$ to get the total width $s+2r$ (similar to the previous case). However, the idea here is to sweep and select $N$ with the largest total width less than the width constraint $w$. The swept values of $N$ vary from $1$ a simple maximum upper-bound estimate taken from the equation for $s$ (which must be positive; thus, positive in its numerator taking $r=0$):