Is there an algorithm to determine what is the maximal determinant you can get just by putting $1$ or $-1$ in a square matrix?
For example in a $3\times3$ matrix:
$$ \begin{bmatrix}1 && -1 && 1\\ 1 && 1 && -1 \\ -1 && -1 && 1\end{bmatrix} $$
Or do you have to go and bruteforce every possibility?
Hadarmard's inequality $$ |\det(A)|^{2} \le \prod_{j=1}^{n}\sum_{k=1}^{n} |a_{jk}|^{2} $$ applied to the case $a_{jk} = \pm 1$ yields $|\det(A)| \le n^{n/2}$. In the $3\times3$ case, this gives $\sqrt{27} \approx 5.19$. If $$ A = \begin{pmatrix} 1 & -1 & -1 \\ 1 & 1 & 1 \\ -1 & -1 & 1 \end{pmatrix} $$ then $\det(A) = 4$, so this inequality isn't a terrible overestimate. No help if you need the exact value though.