$\bullet$ A vector $X=(x_1,\cdots, x_m)$ is less then vector $Y=(y_1,\cdots,y_m)$ when $x_i\leq y_i$, for each $i=1,\cdots, m$, and for at least one $j$, we have $x_j<y_j$.
$\bullet$ A vector $X\in \Pi$ is a minimal vector when there is no any vector, say $Y\in \Pi$, such that $Y<X$.
To be more clear, consider the following example.
Example 1. Assuming $\Pi=\{$(3, 4, 6, 1), (2, 3, 5, 2), (3, 3, 7, 1), (3, 4, 5, 3)$\}$, we have
$\Pi_{min}=\{$(3, 4, 6, 1), (2, 3, 5, 2), (3, 3, 7, 1)$\}$.
Now, assume that a set of m-tuple vectors, say $\Pi$, is given, and we need to find the set of all the minimal vectors in $\Pi$, i.e., $\Pi_{min}$. A vivid approach is to compare all the vectors in $\Pi$ and remove the non-minimal ones. Assuming k as the number of vectors in $\Pi$, since each vector is m-tuple, the time complexity of comparative technique is $O(mk^2)$.
I would like to know if there is a more efficient method for this problem. Actually, I need an approach with time complexity less than $O(mk^2)$.
I highly appreciate your taking time to read my question and to give your answers.