Draw a hollow square of # with given width

Question

I got this challenge from Codingame and am curious about better solutions than mine:

Given a width via standard input draw a hollow square of '#' in given width and length.

Example:

5 results in

#####
#   #
#   #
#   #
#####

I used python to solve this so i am particulary interested in other python code. But please feel free to post your solution in any language you want.

Answer 1

Pyke, 11 bytes

ttDd*n+*.X#

Try it here!

ttDd*n+*    - a square of spaces n-2*n-2 big
        .X# - surround in `#`

Answer 2

Charcoal, 6 bytes

Code:

ＮβＢββ#

Explanation:

Ｎβ        # Get input from the command line and store into β
   Ｂ      # Draw a hollow box with...
     β     #  Width β
      β    #  Height β
       #   #  Filled with the character '#'
           # Implicitly output the box

Try it online!

Answer 3

MATL, 12 bytes

:G\1>&*~35*c

Try it online!

Explanation

:     % Input n implicitly. Push range [1 2 ... n]
      % STACK: [1 2 3 4 5]
G     % Push n again
      % STACK: [1 2 3 4 5], 5
\     % Modulo
      % STACK: [1 2 3 4 0]
1>    % Does each entry exceed 1?
      % STACK: [0 1 1 1 0]
&*    % Matrix with all pair-wise products
      % STACK: [0 0 0 0 0;
                0 1 1 1 0;
                0 1 1 1 0;
                0 1 1 1 0;
                0 0 0 0 0]
~     % Negate
      % STACK: [1 1 1 1 1;
                1 0 0 0 1;
                1 0 0 0 1;
                1 0 0 0 1;
                1 1 1 1 1]
35*   % Multiply by 35
      % STACK: [35 35 35 35 35;
                35  0  0  0 35;
                35  0  0  0 35;
                35  0  0  0 35;
                35 35 35 35 35]
c     % Convert to char. 0 is interpreted as space. Display implicitly
      % STACK: ['#####';
                '#   #';
                '#   #';
                '#   #';
                '#####']

Answer 4

Python 2, 62 54 bytes

f=lambda n:'#'*n+'\n#%s#'%(' '*(n-2))*(n-2)+'\n'+'#'*n

Returns #\n# when the input is 1

55 Bytes version that prints

def f(n):a=n-2;print'#'*n,'\n#%s#'%(' '*a)*a,'\n'+'#'*n

62 Bytes version that works for any input:

f=lambda n:'#'*n+'\n#%s#'%(' '*(n-2))*(n-2)+('\n'+'#'*n)*(n>1)

Answer 5

Jolf, 8 bytes

,ajj"###
,ajj      draw a box with height (input) and width (input)
    "###  with a hash border

Answer 6

COW, 426 405 348 330 bytes

MoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMMMmoOMMMMoOMoOMoOMoOMoOMoOMoOMoOMoO
MoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMMMmoOMMMMoOMoOMoOmoOoomMMM
moOMMMMOOmOomOoMoomoOmoOMOomoomOoMMMmoOMMMMOoMOoMOOmOomOomOomOoMoo
moOmoOMoomoOMMMmoOmoOMMMMOoMOoMOOmOomOomOomOoMoomoOmoOmoOmoOMOomoo
mOomOomOoMoomoOmoOMOomoomOomOomOomOoMoomoOmoOmoOMOOmOoMoomoOMOomoo

Try it online! Change the number in the second line to any number to change the output.

The COW interpreter that I'm using here was written in Perl (and is newer than this challenge), but you can still get the same result by inputting the code here.

Explanation

; Note: [n] means "value stored in the nth block of memory".
MoOMoOMoOMoOMoOMoOMoOMoOMoOMoO                                                  ;Stores 10 in [0].  10 is the code point for carriage return
MMMmoOMMMMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoO     ;Stores 32 in [1].  32 is the code point for whitespace
MMMmoOMMMMoOMoOMoO                                                              ;Stores 35 in [2].  35 is the code point for #
moOoom                                                                          ;Reads STDIN for an integer, and stores it in [3]
MMMmoOMMM                                                                       ;Copies [3] into [4] 
MOO                                                                             ;Loop as long as [4] is non-zero
    mOomOoMoo                                                                   ;Navigate to [2] and print the character with that code point
    moOmoOMOo                                                                   ;Navigate to [4] and decrement
moo                                                                             ;End loop
mOoMMMmoOMMMMOoMOo                                                              ;Copy [3] into [4] and decrement [4] twice
MOO                                                                             ;Loop as long as [4] is non-zero
    mOomOomOomOoMoo                                                             ;Navigate to [0] and print the character with that code point
    moOmoOMoo                                                                   ;Navigate to [2] and print the character with that code point
    moOMMMmoOmoOMMMMOoMOo                                                       ;Navigate to [3] and copy it into [5], then decrement [5] twice
    MOO                                                                         ;Loop as long as [5] is non-zero
        mOomOomOomOoMoo                                                         ;Navigate to [1] and print the character with that code point
        moOmoOmoOmoOMOo                                                         ;Navigate to [5] and decrement
    moo                                                                         ;End loop
    mOomOomOoMoo                                                                ;Navigate to [2] and print the character with that code point
    moOmoOMOo                                                                   ;Navigate to [4] and decrement
moo                                                                             ;End loop
mOomOomOomOoMoo                                                                 ;Navigate to [0] and print the character with that code point
moOmoOmoO                                                                       ;Navigate to [3]
MOO                                                                             ;Loop as long as [3] is non-zero
    mOoMoo                                                                      ;Navigate to [2] and print the character with that code point
    moOMOo                                                                      ;Navigate to [3] and decrement
moo                                                                             ;End loop

Answer 7

Java 7, 113 112 110 bytes

String c(int n){String r="";for(int i=n,j;i-->0;r+="\n")for(j=0;j<n;r+=i*j<1|n-i<2|n-j++<2?"#":" ");return r;}

1 byte saved thanks to @OlivierGrégoire;
2 bytes saved thanks to @cliffroot.

Derived solution based on my Creating a Crossed Square answer.

Try it here.

Answer 8

PowerShell v2+, 48 47 bytes

param($n)($z='#'*$n--);,("#$(' '*--$n)#")*$n;$z

-1 byte thanks to JohnLBevan

Takes input $n, sets $z as $n hashmarks, with $n post-decremented. Encapsulates that in parens to place a copy on the pipeline. Then uses the comma operator to create an array of pre-decremented $n lines of #,spaces,#. Those are left on the pipeline. Then places $z again on the pipeline. Output via implicit Write-Output at the end introduces a newline between elements, so we get that for free.

Since the OP's code doesn't work for input n <= 1, I took that to mean we don't need to support input 1, either.

Examples

PS C:\Tools\Scripts\golfing> 2..6|%{"$_";.\draw-a-hollow-square.ps1 $_;""}
2
##
##

3
###
# #
###

4
####
#  #
#  #
####

5
#####
#   #
#   #
#   #
#####

6
######
#    #
#    #
#    #
#    #
######

Answer 9

Python 2, 59 58 bytes

n=i=input()
while i:print'#%s#'%((' #'[i%n<2])*(n-2));i-=1

repl.it

Note: An input of 1 produces an output of ##, but a hollow square would never be produced for an input less than 3, so I guess this is fine.

Answer 10

C, 98 bytes

f(n,i){i=n*(n+1);while(i--){putchar(i%(n+1)==n?10:i<n||i>n*n-1||i%(n+1)==0||i%(n+1)==n-1?35:32);}}

Usage:

f(5)

Answer 11

05AB1E, 20 bytes

'#×D¹Íð×'#.ø¹Í×sJ¹ä»

Try it online!

Or 18 bytes if we can ignore 1 <= n:

F„ #N¹<%_è¹Í×'#.ø,

Try it online!

Answer 12

JavaScript, 61 58 bytes

Saved 3 bytes thanks to @lmis!

n=>(b='#'[r='repeat'](n))+`
#${' '[r](n-=2)}#`[r](n)+`
`+b

(Doesn't handle 0 or 1)

For 13 extra bytes (at 71 bytes), you can!

n=>n?n-1?(b='#'[r='repeat'](n))+`
#${' '[r](n-=2)}#`[r](n)+`
`+b:'#':''

These solutions are fairly simple: they do a lot of storage to not repeat themselves to save a few bytes. Unminified without the variablsm it would look like:

n => // Anonymous function definition (Param `n` is the size)
    '#'.repeat(n) +      // # `n` times to form the top
    `
#${' '.repeat(n - 2)}#`  // Followed by a newline followed by a hash and `n` - 2 spaces and
                         // another hash to make one of the middle lines
    .repeat(n - 2) +     // The above middle lines repeated `n` - 2 times
    '#'.repeat(n)        // Followed by the top line again

Try it!

<script type="text/babel">var f=n=>n?n-1?(b='#'[r='repeat'](n))+`\n#${' '[r](n-=2)}#`[r](n)+`\n`+b:'#':'',b,r;function c(){document.getElementById('pre').textContent = f(+document.getElementById('input').value);}</script><input id="input" onkeydown="c();" onkeyup="c();" onchange="c();" onclick="c();" placeholder="Size"><pre id="pre"></pre>

Answer 13

WinDbg, 206 200 182 170 bytes

.if@$t0{r$t3=2000000;f@$t3 L@$t0 23;f2*@$t3 L@$t0 20;eb2*@$t3 23;eb2*@$t3+@$t0-1 23;da@$t3 L@$t0;j1<@$t0'.for(r$t1=@$t0-2;@$t1;r$t1=@$t1-1){da2*@$t3 L@$t0};da@$t3 L@$t0'}

-6 bytes from removing parens from .if and using j instead of second .if

-18 bytes by using f instead of a .for to construct the strings.

-12 bytes by not NULL-terminating strings, instead passing length to da

Input is passed in through the pseudo-register $t0 (eg r $t0 = 5; {above-code}).

Explanation:

.if @$t0                                                *Verify width($t0) at least 1 
{                                                       *(registers have unsigned values) 
    r $t3 = 2000000;                                    *Set $t3 to address where the 
                                                        *string will be constructed
    f @$t3 L@$t0 23;                                    *Put width($t0) '#' at 2000000($t3)
    f 2 * @$t3 L@$t0 20;                                *Put width($t0) ' ' at 4000000(2*$t3)
    eb 2 * @$t3 23;                                     *Put '#' on left of ' ' string
    eb 2 * @$t3 + @$t0 - 1 23;                          *Put '#' on right of ' ' string
    da @$t3 L@$t0;                                      *Print the top of the box
    j 1 < @$t0                                          *If width($t1) at least 2
    '
        .for (r $t1 = @$t0 - 2; @$t1; r $t1 = @$t1 - 1) *Loop width($t0)-2 times to...
        {
            da 2 * @$t3 L@$t0                           *...print the sides of the box
        };
        da @$t3 L@$t0                                   *Print the bottom of the box
    '
}

Sample output:

0:000> r$t0=0
0:000> .if@$t0{r$t3=2000000;f@$t3 L@$t0 23;f2*@$t3 L@$t0 20;eb2*@$t3 23;eb2*@$t3+@$t0-1 23;da@$t3 L@$t0;j1<@$t0'.for(r$t1=@$t0-2;@$t1;r$t1=@$t1-1){da2*@$t3 L@$t0};da@$t3 L@$t0'}

0:000> r$t0=1
0:000> .if@$t0{r$t3=2000000;f@$t3 L@$t0 23;f2*@$t3 L@$t0 20;eb2*@$t3 23;eb2*@$t3+@$t0-1 23;da@$t3 L@$t0;j1<@$t0'.for(r$t1=@$t0-2;@$t1;r$t1=@$t1-1){da2*@$t3 L@$t0};da@$t3 L@$t0'}
Filled 0x1 bytes
Filled 0x1 bytes
02000000  "#"

0:000> r$t0=2
0:000> .if@$t0{r$t3=2000000;f@$t3 L@$t0 23;f2*@$t3 L@$t0 20;eb2*@$t3 23;eb2*@$t3+@$t0-1 23;da@$t3 L@$t0;j1<@$t0'.for(r$t1=@$t0-2;@$t1;r$t1=@$t1-1){da2*@$t3 L@$t0};da@$t3 L@$t0'}
Filled 0x2 bytes
Filled 0x2 bytes
02000000  "##"
02000000  "##"

0:000> r$t0=5
0:000> .if@$t0{r$t3=2000000;f@$t3 L@$t0 23;f2*@$t3 L@$t0 20;eb2*@$t3 23;eb2*@$t3+@$t0-1 23;da@$t3 L@$t0;j1<@$t0'.for(r$t1=@$t0-2;@$t1;r$t1=@$t1-1){da2*@$t3 L@$t0};da@$t3 L@$t0'}
Filled 0x5 bytes
Filled 0x5 bytes
02000000  "#####"
04000000  "#   #"
04000000  "#   #"
04000000  "#   #"
02000000  "#####"

Answer 14

Python, 109 bytes

n=int(input())
for x in range(n):
 r=list(' '*n);r[0]=r[-1]='#'
 if x%(n-1)==0:r='#'*n
 print("".join(r))

Answer 15

C, 83 82 80 78 77 Bytes

i,j;f(n){for(i=n;i--;puts(""))for(j=n;j--;putchar(i*j&&i^n-1&&j^n-1?32:35));}

Sneak in a multiply and save a byte...

i,j;f(n){for(i=n;i--;puts(""))for(j=n;j--;putchar(i&&j&&i^n-1&&j^n-1?32:35));}

Also count down j and save a few more...

i,j;f(n){for(i=n;i--;puts(""))for(j=0;j++<n;putchar(i&&j^1&&i^n-1&&j^n?32:35));}

Count down i from n to zero and save a few bytes...

i,j;f(n){for(i=0;i++<n;puts(""))for(j=0;j++<n;putchar(i^1&&j^1&&i^n&&j^n?32:35));}

A bit easier to understand and 1 byte more

i,j;f(n){for(i=0;i++<n;puts(""))for(j=0;j++<n;putchar(i==1|i==n|j==1|j==n?35:32));}

Answer 16

Ruby, 39 bytes

->n{puts a=?#*n,[?#+' '*(n-=2)+?#]*n,a}

Turns out to be shorter this way than all the fancy stuff I was trying. Be advised that this doesn't handle 0 or 1 at all.

Answer 17

Python 2, 50 bytes

m=input()-2
for c in'#'+' '*m+'#':print'#'+m*c+'#'

Works for n>=2. Prints each line with a pound sign, n-2 of the appropriate symbol, then another pound sign.

Aliasing the pound symbol gives same length:

m=input()-2;p='#'
for c in p+' '*m+p:print p+m*c+p

Other attempts:

lambda n:'#'*n+('\n#'+' '*(n-2)+'#')*(n-2)+'\n'+'#'*n

lambda n:'#'*n+'\n#%s#'%((n-2)*' ')*(n-2)+'\n'+'#'*n

lambda n:'\n'.join(['#'*n]+['#'+' '*(n-2)+'#']*(n-2)+['#'*n])

n=input();s='#'+' '*(n-2)+'#'
for c in s:print[s,'#'*n][c>' ']

s='##'+' #'*(input()-2)+'##'
for c in s[::2]:print s[c>' '::2]

s='#'+' '*(input()-2)+'#'
for c in s:print s.replace(' ',c)

Answer 18

Haskell, 49 bytes

n%b='#':(b<$[3..n])++"#\n"
f n=(n%)=<<init(n%' ')

Works for n>=2. Defines the operation of sandwiching a character between # for an n-character newline-terminated string, then applies it twice to make a 2D grid.

Call like:

*Main> putStrLn$ f 5
#####
#   #
#   #
#   #
#####

Answer 19

Groovy, 51 50 bytes

{n->a="*"*n+"\n";n-=2;print(a+"*${' '*n}*\n"*n+a)}

Answer 20

PHP, 81 69 bytes

for($n=-1+$i=$argv[1];$i--;)echo str_pad("#",$n," #"[$i%$n<1]),"#\n";

Run with -r; provide input as argument.

Throws a DivisionByZeroError for input=1.

Answer 21

R, 68 70 bytes

Works for n > 1. Thanks to @Billywob for a couple of bytes swapping out the array for a matrix.

cat(rbind(b<-'#',cbind(b,matrix(' ',n<-scan()-2,n),b),b,'
'),sep='')

Uses rbind and cbind to put rows and columns of #'s around an n-2 square matrix of spaces. Newlines are bound to the rows as well. The newline in the source is significant. Input is from STDIN

Answer 22

Common Lisp, 150 130 bytes

-20 thanks to @Cyoce and @AlexL.

(defun s(v)(format t"~v,,,vA~%"v #\# #\#)(dotimes(h(- v 2))(format t"~v,,,vA~A~%"(- v 1)#\  #\# #\#))(format t"~v,,,vA"v #\# #\#))

Usage:

* (s 5)
#####
#   #
#   #
#   #
#####

Basically uses format twice for the top and bottom and a loop for the rows in between. The format call for the top and bottom outputs a line starting with # and padded to the appropriate width with #s. The format call for the rows in between works similarly, except the padding is spaces and a # gets printed at the end of the line.

Note: I'm rather new to Lisp and expect to have a lot of room for improvement on this.

Answer 23

Python 2, 92 bytes

def f(j):r=range(j);print'\n'.join(''.join(' #'[x*y<1or max(x,y)==j-1]for y in r)for x in r)

Answer 24

ForceLang, 316 314 bytes

def s set
s n io.readnum()
s g goto
def w io.write "#"
s W io.writeln
if n=0
g 3
if n=1
g 4
label 0
if i=n
g 1
s i 1+i
w
g 0
label 1
s i 0
s m n+-2
W()
label a
if i=m
g b
s i 1+i
s j 0
w
label c
if j=m
g d
s j 1+j
io.write " "
g c
label d
W "#"
g a
label b
s i 0
label 2
if i=n
g 3
s i 1+i
w
g 2
label 4
w
label 3

Answer 25

T-SQL 351 bytes

This is so wrong... but anyway:

DECLARE @ INT=1,@n INT=10,@a VARCHAR(MAX),@b VARCHAR(MAX),@c CHAR='#',@s VARCHAR(MAX)=''WHILE @<=@n BEGIN SET @a=concat(@a,@c)SET @b=concat(@b,(SELECT CASE WHEN @=1THEN @c WHEN @=@n THEN @c ELSE' 'END))SET @=@+1 END SET @=1 WHILE @<=@n BEGIN SET @s=@s+(SELECT CASE WHEN @=1THEN @a WHEN @=@n THEN CHAR(13)+@a ELSE CHAR(13)+@b END)SET @=@+1 END PRINT @s

Ungolfed:

DECLARE     @ INT=1,
            @n INT=10, 
            @a VARCHAR(MAX),
            @b VARCHAR(MAX),
            @c CHAR='#',
            @s VARCHAR(MAX)=''
WHILE @<=@n
BEGIN
    SET @a=concat(@a,@c)
    SET @b=concat(@b,(SELECT CASE WHEN @=1THEN @c WHEN @=@n THEN @c ELSE' 'END))
    SET @=@+1;
END
SET @=1
WHILE @<=@n
BEGIN
    SET @s=@s+
    (SELECT 
        CASE 
            WHEN @=1THEN @a
            WHEN @=@n THEN CHAR(13)+@a
            ELSE CHAR(13)+@b 
        END
    )
    SET @=@+1
END
PRINT @s

Answer 26

BrainF***, 152 bytes

+++++[>++>+++++++<<-],[>>>+>+>+>+<<<<<<-]>>>>[-<<.>>]++++[>>>++++++++<<<-]>--<<[<<.>.>>>[>>.<<-]<<<.>->>>[<+<+>>-]<<[>>+<<-]>--<<]<<.>>>>>[<<<<.>>>>-]

Hopefully this can be golfed a bit more.

Explanation:

+++++[>++>+++++++<<-]           Initializes to [0, 10, 35]. 10 is the charcode for a newline, 35 is the charcode for '#'
,                               Gets the input (as an ASCII char, which is converted to an integer)
[>>>+>+>+>+<<<<<<-]             Copies the input to 4 different spaces
>>>>[-<<.>>]                    Prints the appropriate number of '#'
++++[>>>++++++++<<<-]           Puts 32 in the next available space (32 is the charcode of a space)
>--<<--                         Subtracts 2 from one of the input values and the value on which to loop
[                               
  <<.>.>>>                      Print a newline and a '#'
  [>>.<<-]                      Print n - 2 spaces
  <<<.>->>>                     Print another '#'
  [<+<+>>-]<<[>>+<<-]>--<<      Copy n to two other spaces, then copy it back
]                               
<<.>>>>>                        Print a newline
[<<<<.>>>>-]                    Print the appropriate number of '#'

Try it here (visualizer) or here (repl)

Answer 27

Ruby, 66 bytes

n=gets.to_i
puts ?#*n
if n>1
puts [?#+' '*(n-2)+?#]*(n-2),?#*n end

Answer 28

Common Lisp, SBCL, 64 63 bytes

(format t"#~v@{#~}#
~@*~v@{~@*#~v@t#
~}~@*#~v{#~}#"(-(read)2)1)

For n=1 will give:

##
##

but I guess it's ok, since author's solution doesn't seem to work for n=1 (tested here but maybe it's because it's other Python version?)

Other than that it should work fine.

Explanation

format  ;printing function
~v@{#~} ;loop n times, where n equals current argument
        ;printing "#" - gives n "#"s
~@*     ;go back to first argument - reuse (-(read)2)
~v@t    ;add n spaces where n is current argument

Answer 29

Charcoal, 3 bytes (non-competing)

ＢＮ#

Try it online!

Answer 30

Python 2, 60 65 69 62 61 bytes

x=input();h='#';print h*x+'\n'+(h+' '*(x-2)+h+'\n')*(x-2)+h*x

My first attempt at this problem, I will golf it down hopefully.

+5 bytes I forgot to put newlines between the lines :(
+4 bytes I forgot to add 2 to the last x :(
-7 bytes If I don't subtract 2 from the input at the beginning and subtract it where it would make sense to do so, I can actually save 7 bytes :)
-1 byte Thanks to @Cyoce for helping me save 1 byte by assigning '#' to a variable (which I called h because that's what he was thinking in the comments :P)

Still 1 byte longer than before... :(

Answer 31

Haskell, 67 bytes

l#n=l<$[1..n]
f n=unlines$'#'#n:('#':' '#(n-2)++"#")#(n-2)++['#'#n]

Usage example:

Prelude> putStrLn $ f 4
####
#  #
#  #
####

How it works:

l#n=l<$[1..n]                      -- helper function that makes n copies of l

   '#'#n                           -- make a string of n copies of #, followed by
                        #(n-2)     -- n-2 copies of
     '#':' '#(n-2)++"#"            -- # followed by n-2 times spaces, followed by #
                           ['#'#n] -- and a final string with n copies of #
unlines                            -- join with newlines in-between

Answer 32

Jelly, 13, bytes

,þ%µỊṀ€€ị⁾# Y

TryItOnline! or try 0 to 15

How?

,þ%µỊṀ€€ị⁾# Y - Main link: n
 þ            - outer product with
,             -    pair:   [[[1,1],[2,1],...,[n,1]],[[1,2],[2,2],...,[n,2]], ... ,[[1,n],[2,n],...,[n,n]]]
  %           - mod n:     [[[1,1],[2,1],...,[0,1]],[[1,2],[2,2],...,[0,2]], ... ,[[1,0],[2,0],...,[0,0]]]
   µ          - monadic chain separation
    Ị         - abs(z)<=1: [[[1,1],[0,1],...,[1,1]],[[1,0],[0,0],...,[1,0]], ... ,[[1,1],[0,1],...,[1,1]]]
      €€      - for each for each
     Ṁ        - maximum:   [[1,    1,    ...,1],    [1,    0,    ..., 1],    ... ,[1,    1,    ..., 1]   ]
        ị     - index into (1 based)
         ⁾#   - "# ":      ["##...#","# ...#", ...,"##...#"]
           Y  - join with line feeds

Answer 33

Racket 113 bytes

(let*((d display)(g(λ()(for((i n))(d"#")))))(g)(d"\n")(for((i(- n 2)))(d"#")(for((i(- n 2)))(d" "))(d"#\n"))(g))

Ungolfed:

(define (f n)
  (let* ((d display)
         (g (λ () 
              (for ((i n))
                (d "#"))
              (d "\n"))))
    (g)
    (for ((i (- n 2)))
      (d "#")
      (for ((i (- n 2)))
        (d " ") )
      (d "#\n"))
    (g)))

Testing:

(f 5)

Output:

#####
#   #
#   #
#   #
#####

Answer 34

Pip, 16 bytes

15 bytes of code, +1 for -n flag.

(Y_Xa-2WR'#s)My

Works for input >= 2. Try it online!

Explanation of somewhat ungolfed version

First, we define a function y that takes a string argument, repeats it a-2 times (where a is the first command-line input), and wraps the result in #.

Y _ X a-2 WR '#
  _              Identity function
    X a-2        String-repeated by a-2
          WR '#  Wrapped in #
Y                Yank the resulting function into y

Next, we apply this function twice--once normally, then again with map--to obtain the square as a list of strings:

y M (y s)
    (y s)  Call function y with s (preinitialized to " ") as argument
y M        Map y to each character of the resulting string

For input of 4, (y s) results in "# #" and y M (y s) in ["####"; "# #"; "# #"; "####"]. This latter value is then printed, with the -n flag causing it to be newline-separated.

Golfing tricks

To get from the ungolfed to the golfed version:

• Remove spaces.
• Y is an operator, which means we can use it in an expression. Instead of Y... followed by (ys), we can just do (Y...s).
• The problem is, we have to yank the function before we reference it again as y; so yM(Y_Xa-2WR'#s) won't work. Solution: swap the operands of the Map operator. As long as one of them is a function and the other is an iterable type, it doesn't matter what order they come in.

Answer 35

C#, 154 152 bytes

Golfed:

void F(int n){Console.Write($"{new string('#',n)}\n");for(int i=2;i<n;i++)Console.Write($"#{new string(' ',n-2)}#\n");Console.Write(new string('#',n));}

Ungolfed:

    void F(int n)
    {
        Console.Write($"{new string('#', n)}\n");

        for (int i = 2; i < n; i++)
            Console.Write($"#{new string(' ', n - 2)}#\n");

        Console.Write(new string('#', n));
    }

EDIT1: Loop range optimization.

Answer 36

SpecBAS - 57 bytes

1 INPUT n: a$="#"*n,n-=2,b$="#"+" "*n+"#"#13: ?a$'b$*n;a$

? is shorthand for PRINT, #13 is carriage return which can be tacked on to the end of a string without needing a + to join them.

The apostrophe moves print cursor down one line.

Answer 37

Stuck, 29 27 Bytes

Pretty darn long for a "golfing" language, but I have forgotten how a lot of it works :P

i_2-_u'#*N+_'#' u*'#N+++u*u

Explanation:

i_2-_u                           # take input and triplicate, subtracting 2 (5 -> [3,3,5])
      '#*N+_                     # create the top and bottom rows
            '#' u*'#N+++u*       # create input - 2 copies of middle rows
                          u      # rotate left 1 to get correct order, implicit output

Answer 38

Lithp, 117 bytes

Line split in two for readability:

#N::((var X (repeat "#" N))(print X)(each (seq 3 N) (scope #X::((print (+ "#" 
     (repeat " " (- N 2)) "#")))))(print X))

Sample usage:

% square.lithp
(
    (import "lists")
    (def s #N::((var X (repeat "#" N))(print X)(each (seq 3 N) (scope #X::((print (+ "#" (repeat " " (- N 2)) "#")))))(print X)))
    (s 10)
)

Output:
$ ./run square.lithp
##########
#        #
#        #
#        #
#        #
#        #
#        #
#        #
#        #
##########

Answer 39

dc, 83 bytes

?sd0[1+[#]Pdld>k]sklkx0[1+[
]P0[#]P[1+[ ]Pdld2->x]sxlxxk[#]Pdld2->r]srld2<r[
]P0lkx

This is mostly straightforward. It's kinda late so I won't write up an explanation now, but I might edit one in later.

(Note: I'm new to dc so there are probably obvious improvements to be made here.)