Let P be a set of n points in the plane. We compute the value of $$\theta \in [0,2\pi )$$ for which the rectilinear convex hull of P, denoted by $$\mathcal {RH}_{P}({\theta })$$ , has minimum (or maximum) area in optimal $$O(n\log n)$$ time and O(n) space, improving the previous $$O(n^2)$$ bound. Let $$\mathcal {O}$$ be a set of k lines through the origin sorted by slope and let $$\alpha _i$$ be the sizes of the 2k angles defined by pairs of two consecutive lines, $$i=1, \ldots , 2k$$ . Let $$\Theta _{i}=\pi -\alpha _i$$ and $$\Theta =\min \{\Theta _i :i=1,\ldots ,2k\}$$ . We obtain: (1) Given a set $$\mathcal {O}$$ such that $$\Theta \ge \frac{\pi }{2}$$ , we provide an algorithm to compute the $$\mathcal {O}$$ -convex hull of P in optimal $$O(n\log n)$$ time and O(n) space; If $$\Theta < \frac{\pi }{2}$$ , the time and space complexities are $$O(\frac{n}{\Theta }\log n)$$ and $$O(\frac{n}{\Theta })$$ respectively. (2) Given a set $$\mathcal {O}$$ such that $$\Theta \ge \frac{\pi }{2}$$ , we compute and maintain the boundary of the $${\mathcal {O}}_{\theta }$$ -convex hull of P for $$\theta \in [0,2\pi )$$ in $$O(kn\log n)$$ time and O(kn) space, or if $$\Theta < \frac{\pi }{2}$$ , in $$O(k\frac{n}{\Theta }\log n)$$ time and $$O(k\frac{n}{\Theta })$$ space. (3) Finally, given a set $$\mathcal {O}$$ such that $$\Theta \ge \frac{\pi }{2}$$ , we compute, in $$O(kn\log n)$$ time and O(kn) space, the angle $$\theta \in [0,2\pi )$$ such that the $${\mathcal {O}}_{\theta }$$ -convex hull of P has minimum (or maximum) area over all $$\theta \in [0,2\pi )$$ .
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