What's being surrounded
Perl, 144 138 132 129 128 127 126 124 bytes
Includes +2 for -p0
The code assumes \0 is not a valid input character (at least inside the #).
Run with the input on STDIN:
surround.pl < surround.txt
surround.pl:
#!/usr/bin/perl -p0
/^#[^#\0]/m&&s/^|[^#\n\0]\0/\0\0/mg,s%.%s/.(.*)/$+\0/g;/#/&&reverse"\n",/^./mg%seg until$?++<$$_++;y/\0/#/;s/^#*\n|#+$|^#//mg;y;#;
The code works as is, but replace the \0 and \n by their literal versions for the claimed score. Notice there is a space at the end of the line. The code loops way too many times, so you may have to wait 30 seconds or so for output.
Explanation
I am going to do a floodfill with \0 stopped by # from the outside in the orthogonal directions. After that I will slice off the # sides and replace all that is left by spaces. To avoid having to handle all directions in the floodfill I will repeatedly rotate the target area and only floodfill from right to left
/^#[^#\0]/m The rotation is written such that it slices
off the first column. That is ok unless the
first column contains a # that is followed by
something that could be the inside. There is
no newline inside the [] because short lines
will get extended during the rotation and
the character following the # will end
up as a \0 and match in a later round
&&s/^|[^#\n\0]\0/\0\0/mg In case the # could be an interior border I
will add two columns of \0's in front. One
will be a sacrifice for the rotation, the
other column will end up at the end of the area
after two rotations and function as seed for the
floodfill. This regex also does one step of
the floodfill from the back to the front.
After a certain number of loops we are certain
to get to a first column that must not be
dropped so at some point the last column is
guaranteed to consist of only \0. And we only need
to fill backward since the rotations will make
any direction backward at some point
s%.% process column %seg I will replace each character (including \n)
in the string by the next column in reversed
order or an empty string if there are no more
interesting columns. This is therefore a right
rotation. There are less columns than
characters so this loop is long enough
s%.%s/.(.*)/$+\0/g Remove the next (now first) character from each
row (so remove the column). Because the
original area is not necessarily a rectangle
add a \0 at the end of the row so we won't run
out out of columns (this would cause shorter
rows to have no entry in the new rotated row)
This will not do anything for empty lines so
they DO get squeezed out. But that is not a
problem since the problem statement says there
will be only one # shape so any empty lines
are can be safely dropped (this would not be
so if there could be multiple # shapes because
that could create a new surrounded area
/#/ Check if any of the remaining columns still
has a #. If not all remaining columns are on
the outside and can be dropped
&&reverse"\n",/^./mg Collect the column and add a \n to its reverse
until$?++<$$_++ Keep doing this until we get to a multiple of
65536 rotations when $? waraps back around to 0
(this is a multiple of 4 so the area is left
unrotated) and an area we have seen before
($$_ >= 1)
(so all slicing and flood filling is finished)
$_ having been seen in a previous rotations is
not a problem (though rather tricky to prove)
At this point e.g.
AB##J
E####GK
F# M #L
# N#O
P####
will have been replaced by:
0000000
0####00
0# M #0
# N#0
0####00
Basically all columns and rows that are not directly bordering the inside have been sliced off. Any outside characters left has been replaced by \0. At the top and right there is an extra layer of \0. So all that is left is cleanup:
y/\0/#/ Replace any outside that is left by #
s/^#*\n|#+$|^#//mg Removes the first two and last line (the only
lines that can consist of purely #)
Removes any trailing #
Removes the first column of #
y;#; \n; Replace any remaining # by space since they
are needed to fill the concave parts
The final \n; is not written since it is implicit
in the -p loop
Javascript, 485 464 427 417 396 390 bytes
s='indexOf';k='lastIndexOf';h="#";t=b=>b[0].map((x,i)=>b.map(x=>x[i]));i=>{m=i.split`
`;for(h of m){m[m[s](h)]=h.split``;}for(y=0;y<m.length;y++){for(z=x=0;x<m[y].length;x++){if(m[y][x]==h)break;if(m[y][s](h)<x&&m[y][k](h)>x)z++;q=t(m);if(q[y][s]h)<x&&m[y][k](h)>x)z++;if(z>2)m[y][x]=h}}for(p of m){v=p.join``.match(/\S/);e=v?p.join``:'';m[m[s](p)]=e;}m=m.join`
`;return m.replace(#/g," ")}Yes. I tried. And, although I am at 485 bytes, I am winning because no-one else felt like answering this question. So, hah!
And also, I am well aware that I could golf this loads, I just am tired at the moment... well now I'm at 396 Thanks to Conor for most of the golfing... :D