Notice that if somewhere in the table there is a diamond shape whose edge is made out of #, then its interior is also a diamond shape, but its edge is from . instead of #. Therefore, it is enough to iterate over the table and using BFS or DFS, find the maximal connected regions made from . and for each such region, check if it has a diamond shape.

Let's now look at all cells $(i,j)$ for which $i+j=k$, for some fixed value of $k$. It's easy to see that those cells will form a diagonal in the table going in the direction from bottom-left to top-right. Also, smaller values of $k$ correspond to diagonals closer to the top-left corner of the table, while larger values of $k$ correspond to diagonals closer to the bottom-right corner of the table. In a similar way, the set of cells for which $i-j=k$ forms a diagonal, but in the other direction. This means that if we are given fixed numbers $a$, $b$, $c$ and $d$ such that $a\le b$ and $c\le d$, then the set of all cells $(i,j)$ for which $a\le i+j\le b$ and $c\le i-j\le d$ actually forms a tilted rectangle. In case $b-a=d-c$, it is actually a tilted square, i.e. a diamond.

Therefore, to check if some region is a diamond or not, we keep track of the following: the number of visited cells (cnt), and the minimum and maximum diagonal in both directions ($a=min(i+j)$, $b=max(i+j)$, $c=min(i-j)$, $d=max(i-j)$). What remains is to check whether $b-a=d-c$ and to see if the number of visited cells matches the number of cells that should be in the diamond of this size ($b-a$). In doing so, we need to calculate the number of . in a diamond of size $n$, which turns out to be $n^2+(n-1{)}^2$.