Fill in the lakes

Question

Given the topography of land in ASCII picture format, figure out where lakes would go and fill them in. Assume an infinite amount of rain.

example

input

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

output

         #               
         ##              
      #@####             
#@@@@#########           
##@@###########          
##@#############@@@####  
##@##############@@##### 
#################@#######
#########################
#########################

The input will contain only spaces and # marks. Each line will be the same length. The output should be the identical # pattern with spaces where water would accumulate filled in with @ marks.

The bottom input row will always be all # marks. There will be no holes or overhangs in the land. Shortest code wins.

Answer 1

sed -r, 27 24 (27 with -r)

24 (27):

:;s/(#|@) ( *#)/\1@\2/;t

27 (30):

:e;s/([#@]) ( *#)/\1@\2/;te

Competes with the better of the two perl solutions

Answer 2

Perl, 25

s/# +#/$_=$&;y| |@|;$_/ge

Answer 3

Perl (>= v5.9.5), 24 chars

Run with perl -p:

1while s/#.*\K (?=\S)/@/

This requires Perl 5.9.5 or later to use the special escape \K.

Answer 4

Windows PowerShell, 36 74 138

$input-replace'(?<!^ *) (?! *$)','@'

Answer 5

Retina, 10 bytes

Retina is (much) newer than this challenge. But this solution is too neat not to post it:

T` `@`#.*#

Try it online.

This is simply a transliteration stage which replaces spaces with @, but the operation is restricted to matches of #.*#, i.e. characters which are surrounded by land on both sides.

Answer 6

05AB1E, 17 16 bytes

|εγć?D¨ð'@:sθJJ,

Try it online.

Explanation:

|            # Take all input-lines as list
 ε           # For each line:
  γ          #  Split the line into chunks of consecutive equal characters
             #   i.e. " ##   # " → [' ','##','   ','#',' ']
   ć         #  Split into head and the rest of the list
             #   i.e. [' ','##','   ','#',' '] → ['##','   ','#',' '] and ' '
    ?        #  Print this head
   D         #  Duplicate the rest of the list
    ¨        #  Remove the last element
             #   i.e. ['##','   ','#',' '] → ['##','   ','#']
     ð'@:    #  Replace every space with a "@"
             #   i.e. ['##','   ','#'] → ['##','@@@','#']
     sθ      #  Swap so the duplicated list is at the top, and take the last item as is
             #   i.e. ['##','   ','#',' '] → ' '
         JJ  #  Join the lists and individual items in the list together to a single string
             #   i.e. ['##','@@@','#'] and ' ' → "##@@@# "
           , #  Print with trailing new-line

Answer 7

Ruby 1.8, 30 characters

#!ruby -p
gsub(/# +#/){$&.tr" ","@"}

If anyone has an idea why this doesn't work in Ruby 1.9 (tested with 1.9.2p0 and 1.9.2p204), even though the documentation says it should work, let me know!

Answer 8

Python, 95 92 bytes

for s in S.split('\n'):b=s.find('#');e=s.rfind('#');print s[:b]+s[b:e].replace(' ','@')+s[e:]

Answer 9

Jelly, 15 bytes

Ỵ”@ẹ”#ṂrṀḟƲ$¦€Y

Try it online!

Answer 10

Pip, 15 bytes

aR:`#.*#`_TRs'@

Takes input as a multiline string via command-line argument: Try it online! (Alternately, specify the -rn flags and change the first a to g, and you can give input via stdin: Try it online!)

Same idea as the Retina answer: replace every match of the regex #.*# with the result of transliterating space to @ in the match. Pip can't match Retina's terseness for a pure regex problem--but it's not everyday that you can tie with Jelly, after all.

Answer 11

Python, 108 106 92 bytes

import re
n=1
while n: S,n=re.subn('# +#',lambda m:'#'+'@'*len(m.group(0)[2:])+'#',S)
print S

Answer 12

Javascript, 107 bytes

var f=function(x){return x.replace(/# +#/g, function(x){return "#"+new Array(x.length-1).join("@")+"#";})};

Ungolfed:

var f = function(x) {
    return x.replace(/# +#/g, function(x){
        return "#" + new Array(x.length - 1).join("@") + "#";
    })
};