# Build me a city

Coders are always trying to flatten arrays into boring 1-dimensional entities and it makes me sad.

Your task is to unflatten an arbitrary string of characters, outputting a lovely city skyscape.

Consider the string: aaabbbbbccqrrssstttttttPPw

It looks much better like this:

            tt
tt
bb        tt
bb        tt
aabb      sstt
aabbcc  rrssttPP
aabbccqqrrssttPPww


(Ok, yes, the letters are duplicated to make it look more city-skyline-ery).

Take an input string, duplicate each subsection of matching characters (not necessarily alphabetic letters) and build me a city!

Shortest code bytes win.

I actually thought I had the requirements nailed, but to answer some questions:

• it must be on the ground
• you can have extra sky if you want (leading blank lines, surrounding blank space) - but not between the buildings
• letters can be reused inside the string (same architecture, different location)
• the letters are assumed to be ASCII, but more flair will be given to those supporting additional encodings (UTF8, etc)
• Can we output the cityscape rotated 90 degrees?
– Okx
Jun 27 '17 at 11:30
• Will characters ever repeat again i.e. aaabbbbaa? Jun 27 '17 at 11:30
• @Okx have you ever seen a city rotated 90 degrees, that would look very silly! ;)
– Tom
Jun 27 '17 at 11:30
• @Tom maybe..?
– Rod
Jun 27 '17 at 11:31
• Welcome on the site! For future challenges, I recommend posting them first in the Sandbox where you can get feedbacks from the community before posting it as a challenge.
– Dada
Jun 27 '17 at 11:32

# 05AB1E, 6 bytes

γ€DζR»


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In a version newer than the challenge, ζ has been added as a replacement fo .Bø

# 05AB1E, 8 bytes

γ€D.BøR»


Explanation:

γ            Convert into a list of consecutive equal elements
€D          Duplicate each element
.B        Squarify; pad each element with spaces so that they are the length of the longest element
ø       Transpose
R      Reverse (otherwise the city would be upside-down)
»     Join by newlines


Try it online!

• Interestingly, Jelly has z⁶ for .Bø...but it also has Œgx'2 for γ€D >_> Jun 27 '17 at 11:46
• γ.BD)ø˜øR» was what I had without looking, €D is way better; I feel like we're both missing the 1-byte solution to inline duplication though. Jun 27 '17 at 20:25
• @MagicOctopusUrn Wait, you solved the challenge without even looking at it?
– Okx
Jun 27 '17 at 20:51
• @Okx Well, it's wise not to look at the answers before since the whole fun of golfing it yourself could be cut off. Jun 29 '17 at 15:24
• @EriktheOutgolfer It was a joke, and what I mean was that he solved it without looking at the content of the challenge.
– Okx
Jun 29 '17 at 16:10

# CJam, 23 bytes

qe::*:__:,:e>f{Se[}zN*


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Explanation:

qe::*:__:,:e>f{Se[}zN* Accepts (multi-line?) input
q                       Take all input
e::*                  Split into groups of equal elements
:_                Duplicate each
_:,:e>          Push maximal length without popping
f{Se[}    Left-pad each to that length with space strings (NOT space chars, although not a problem here)
z   Zip
N* Join with newlines

• Wow, a CJam answer >_> Jun 27 '17 at 12:08

# Jelly, 9 bytes

Œgx'2z⁶ṚY


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Explanation:

Œgx'2z⁶ṚY  Main Link
Œg         Group runs of equal elements
x        Repeat
'              the lists
2                       twice without wrapping
z⁶    Zip (transpose), filling in blanks with spaces
Ṛ   Reverse the whole thing so it's upside-down
Y  Join by newlines

• Could you add an explanation please milord? I can't understand what's happening here :o Jun 27 '17 at 15:02
• @Nathan pastebin.com/DzaBSjd5 Jun 27 '17 at 19:24
• @HyperNeutrino Nice explanation... Jun 27 '17 at 19:25
• Just to be sure, is it correct? :P Jun 27 '17 at 19:33
• @HyperNeutrino Well, that wasn't entirely the intention for ', which was to repeat the lists themselves and not the items inside them, but overall it's good. :) Jun 27 '17 at 19:38

# Python 3, 155136134 132 bytes

-19 bytes thanks to @LeakyNun
-2 bytes thanks to @officialaimm
-1 byte thanks to @Wondercricket

s=input()+'+'
k=' '*len(s)
a=[]
c=b=''
while s:
while c in b:b+=c;c,*s=s
a+=b+k,b+k;b=c
for r in[*zip(*a)][:0:-1]:print(*r,sep='')


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# Python 2, 117 bytes

import re
s=input()
for l in zip(*[x+S*len(s)for x,_ in re.findall(r'((.)\2*)',s)for S in'  '])[::-1]:print''.join(l)


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# Java 8, 412400330324312 319 bytes

-6 bytes thanks to VisualMelon
-12 bytes thanks to Kevin Cruijssen
but +19 bytes because I forgot to include the imports in the byte count.

import java.util.*;x->{Map m=new HashMap(),n;int l=x.length(),i=l,v,y,h=0,d=1;char c,k;for(;i-->0;m.put(c,d=m.get(c)!=null?d+1:1),h=d>h?d:h)c=x.charAt(i);for(y=h;y>0;y--){n=new HashMap(m);for(i=0;i<l;i++)if(n.get(k=x.charAt(i))!=null){v=(int)m.get(k);System.out.print((y>v?"  ":k+""+k)+(i==l-1?"\n":""));n.remove(k);}}}


Try it online!

• Golfing Java and C# (my department) is great fun! Keep at it! Not tested, but I think you can save a few bytes by rejigging the for loops: you can pre-assign i=0, or better, i=l, and count down for(;i-->0;h=d>h?d:h) (and stuff the h= bit in there). The same back-counting will work for the inner loop also. The inner if also has no need for the braces {}. And always be weary of <= or >=, you can turn the ternary around with > and save a byte. Jun 28 '17 at 12:10
• Thanks, I could shave another 6 bytes off the code thanks to your tips. Well, I think I'm going to stay with Java Golfing since I actually like it ;). Jun 28 '17 at 14:59
• Welcome to PPCG! I'm afraid you'll have to increase the byte-count to 329 (+19 bytes due to the required import java.util.*; for Map and HashMap, imports are part of the byte-count; and -1 by removing trailing semi-colon, which isn't part of the byte-count). Jun 29 '17 at 12:57
• Jun 29 '17 at 12:58
• Summary of the changes: HashMap<>HashMap; Map n=,n and n=; m.put(c,d=m.get(c)!=null?d+1:1); inside the for-loop to get rid of the brackets; k=x.charAt(i) inside the if(n.get(k)!=null) to get rid of the semi-colon and for-loop's brackets. Again, welcome and great answer! +1 from me. Also, in case you haven't seen it yet: Tips for golfing in Java and Tips for golfing in <any language> might be interesting to read through. Jun 29 '17 at 13:02

# Japt-R, 1918151312 9 bytes

Includes trailing spaces on each line.

ò¦ Õ®m²ÃÔ


Try it

ò¦ Õ®m²ÃÔ     :Implicit input of string
ò             :Partition by
¦            :  Inequality
Õ          :Transpose
®         :Map
m        :  Map
²       :    Duplicate
Ã      :End map
Ô     :Reverse
:Implicit output joined with newlines


# Mathematica, 150 bytes

(z=Characters[v=#];f=CharacterCounts[v][#]&/@(d=Union@z);Row[Column/@Map[PadLeft[#,Max@f,""]&,Table[Table[d[[i]]<>d[[i]],f[[i]]],{i,Length@d}],{1}]])&


# R, 135 bytes

e=rle(sub('(.)','\\1\\1',strsplit(scan(,''),'')[[1]]));write(sapply(sum(e$l|1):1,function(x)ifelse(e$l>=x,e$v,' ')),'',sum(e$l|1),,'')


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reads from stdin, writes to stdout (with a trailing newline).

### Explanation:

• rle finds the lengths of the streaks of characters, the heights of each tower.
• the sub expression replaces each character with its double (so I didn't have to muck about with setting adjacent indices together)
• sapply returns an array (in this case a matrix):
• sum(e$l|1) is the number of distinct characters; we go from top to bottom • ifelse( ... ) is a vectorized if...else allowing us to build a matrix of towers and double spaces • write writes to console, with a few options to format. # Pyth, 13 bytes j_.tsC*2]*Mr8  Try it online! • 49 minutes... way too slow – Okx Jun 27 '17 at 12:17 # PHP, 138 bytes for(;~$s=&$argn;$s=substr($s,$r++%2*$i))for($i=0;$i<strspn($s,$c=$s[0]);$f[$i++][$r]=$c)a&$f[$i]?:$f[$i]=" ";krsort($f);echo join(" ",$f);


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# MATL, 15 bytes

'(.)\1*'XXtvc!P


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### Explanation

'(.)\1*' % Push string to be used as regexp pattern
XX       % Implicit input. Regexp matching. Pushes row cell array of matching substrings
t        % Duplicate
v        % Concatenate vertically
c        % Convert to char. This reads cells in column-major order (down, then across)
% and produces a 2D char array, right-padding with spaces
!        % Transpose
P        % Flip vertically. Implicitly display


# Charcoal, 40 bytes:

Ａ⟦⟦ω⟧⟧λＦθ¿⁼ι§§λ±¹¦⁰⊞§λ±¹ι⊞λ⟦ι⟧ＦλＦ²↑⁺⪫ιω¶


Try it online! Link is to verbose version of code. I originally tried a simple loop over the input string to print an oblong every time the letter changed, but I switched to this list-building method as it saved 5 bytes. Explanation: The variable l contains a nested list of the input letters. Characters that match the current last list elements get pushed onto the last list otherwise a new sublist is created for that character. It then remains to join the letters in each sublist so that they can be printed vertically twice.

# QuadS, 15 + 1 = 16 bytes

+1 byte for the 1 flag.

⊖⍵
(.)\1*
2/⍪⍵M


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⊖⍵ post-process by flipping upside down

(.)\1* runs of identical characters

2/⍪⍵M duplicate the columnified Match

The 1 flag causes the results to be merged together.

# Ruby, 116 bytes

->s{a=s.scan(/(.)(\1*)/).map{|x,y|[x,y.size+1]}.to_h
m=a.values.max
m.times{|i|puts a.map{|k,v|v+i<m ?'  ':k*2}*''}}


Try it online!

• puts a.map{...} can be replaced with p(a.map{}) Jun 29 '17 at 0:06
• p will output quote characters, so it's not fit here
– Alex
Jun 29 '17 at 5:37
• oh, wow, thanks. You learn everyday - stackoverflow.com/a/1255362/2047418 Jun 29 '17 at 7:45

# C, 259 231 Bytes

## Golfed Code

#define v a[1][i
i,k,l,x,h,w;main(char*s,char**a){for(;v];w+=2*!x,s=v++],h=x>h?x:h)x=(s==v])*(x+1);h++;s=malloc((x=h++*++w+1)+w);memset(s,32,h*w);for(i=k;v];s[x+1]=s[x]=k=v++],x=k==v]?x-w:(h-1)*w+l++*2+3)s[i*w]=10;printf("%s",s);}


## Verbose Code

//Variable Explanations:
//i - increment through argument string, must beinitialized to 0
//k - increment through argument string, must be initialized to 0
//l - record x coordinate in return value, must be initialized to 0
//x - record the actual character position within the return string
//arrheight - the height of the return string
//arrwidth - the width of the return string
//arr - the return string
//argv - the string containing the arguments
#define v argv[1][i

i,k,l,x,arrheight,arrwidth;

main(char*arr,char**argv){
for(;v];                                 //For Length of input
arrwidth+=2*!x,                        //increment width by 2 if this char is not the same as the last
arr=v++],                              //set arr to current char
arrheight=x>arrheight?x:arrheight      //see if x is greater than the largest recorded height
)x=(arr==v])*(x+1);                     //if this character is the same as the last, increment x (using arr to store previous char)
arrheight++;                             //increment height by one since its 0 indexed
arr=malloc((x=arrheight++*++arrwidth+1)+arrwidth); //create a flattened array widthxheight and set x to be the bottom left position
memset(arr,32,arrheight*arrwidth);       //fill array with spaces
for(i=k;v];                              //For Length of input
arr[x+1]=arr[x]=k=v++],                //set x and x+1 positions to the current character, store current character in i
x=k==v]?x-arrwidth:(arrheight-1)*arrwidth+l++*2+3 //if next char is same as current move vertically, else set x to bottom of next column
)arr[i*arrwidth]=10;                     //Add new lines to string at end of width

printf("%s",arr);                        //output string

}


Compiled with GCC, no special Flags

### Edit

Saved 28 bytes thanks to adelphus. His change allowed me to create a define. And I made the while loops into for loops to save 2 bytes each by rearranging the loop. I also fixed an issue where the code would break when the last character in input wasn't singleton. The code will fail if there is only one unique letter but should work in all other cases.

• Nice! But the golfed version doesn't seem to work with arbitrary input for some reason. Removing the final "w" from the sample input appears to lose the q's and repeat the string. Sure it's something small... Jun 28 '17 at 23:01
• also while (i < strlen(argv[1])) can be shortened to while (argv[1][i]) - loop until null character Jun 28 '17 at 23:06
• @adelphus Interesting, I'll try it out tomorrow when I get a chance. I did not test anything other than the given test case (lazy I know). Jun 29 '17 at 0:54
• That actually helped a ton, I was able to fix the problem and reduce by almost 30 bytes! Jun 29 '17 at 12:46

# Factor + grouping.extras sequences.repeating, 94 bytes

[ [ ] group-by values dup longest length '[ _ 32 pad-head ] map 2 repeat flip [ print ] each ]


Modern Factor doesn't require a conversion to string before printing a sequence of code points, so have a screenshot of running the quotation in the listener:

Here's a TIO link with the extra >string:

Try it online!

## Explanation:

It's a quotation (anonymous function) that takes a string as input and outputs to standard output. Assuming "aaabbbbbccqrrssstttttttPPw" is on the data stack when this quotation is called...

• [ ] group-by values Split into groups of equal elements.

Stack: { V{ 97 97 97 } V{ 98 98 98 98 98 } V{ 99 99 } V{ 113 } V{ 114 114 } V{ 115 115 115 } V{ 116 116 116 116 116 116 116 } V{ 80 80 } V{ 119 } }

• dup longest length Keep the 'city' on the stack but also find the length of the longest 'building.'

Stack: ~city~ 7

• '[ _ 32 pad-head ] map Pad the city with whitespace to be a rectangular matrix.

Stack: { V{ 32 32 32 32 97 97 97 } V{ 32 32 98 98 98 98 98 } V{ 32 32 32 32 32 99 99 } V{ 32 32 32 32 32 32 113 } V{ 32 32 32 32 32 114 114 } V{ 32 32 32 32 115 115 115 } V{ 116 116 116 116 116 116 116 } V{ 32 32 32 32 32 80 80 } V{ 32 32 32 32 32 32 119 } }

• 2 repeat Repeat each row twice.

• flip Transpose the matrix.

• [ print ] each Print each row to standard output.

# Pip, 22 bytes

21 bytes of code, +1 for -l flag.

Ya@(.)\1*RV:yWVyZDs


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### Explanation

                       a is 1st cmdline arg; s is space (implicit)
a@(.)\1*            Using regex, create list of runs of same character in a
Y                      Yank that into y variable
yWVy     Weave (interleave) y with itself to duplicate each item
ZDs  Zip to transpose, with a default character of space filling gaps
RV:         Reverse the resulting list (with the compute-and-assign
meta-operator : being abused to lower the precedence)
Auto-print, one sublist per line (implicit, -l flag)


# Haskell, 144 bytes

f s=let x=groupBy(==)s;l=length;m=maximum(map l x)in concatMap(++"\n")$reverse$transpose$concat[[z,z]|z<-(map(\y->y++(replicate(m-(l y))' '))x)]  I'm pretty confident I can do better than this, but this is the best I can come up with for the time being. • Bad news first: you use functions from Data.List which is not in scope by default. You have to either add the import Data.List to the byte count or specify a Haskell environment which does include it by default (e.g. change the language from Haskell to Haskell (lambdabot). -- Some tips: a) use pattern guards to bind variables instead of let and/or declare helper functions directly: l=length;f s|x<-groupBy(==)s,m<-... =concatMap. b) map l x is l<$>x, c) concatMap("++\n" is unlines. d) groupBy(==) is just group. e) concat is id=<<. You use m only once, so inline it
– nimi
Jun 28 '17 at 17:37
• ... f) no need for () around l y, replicate ... ' ' and map ... x. All in all: import Data.List;l=length;f s|x<-group s=unlines$reverse$transpose$id=<<[[z,z]|z<-map(\y->y++replicate(maximum(l<$>x)-l y)' ')x].
– nimi
Jun 28 '17 at 17:38
• groupBy(==)=group, altough I'm not sure whether one is in Prelude and the other isn't. concatMap can be written >>=, and map can be infixed as <$>, and concat[[z,z]|z<-…] might be (replicate 2)=<<… or (\z->[z,z])=<<… Jun 28 '17 at 17:44 • You can shave off one more byte from @Bergi's excellent tip: (\z->[z,z]) is (:)<*>pure, i.e. ...transpose$(:)<*>pure=<<map(\y...)x
– nimi
Jun 28 '17 at 17:59

# Stacked, 42 bytes

[rle toarr[...*chars]map FIX 2*tr rev out]


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# Ruby, 107 bytes

gets.gsub(/(.)(\1*)/,'\0
'*2).split('
').map{|x|x.ljust($_.size).chars}.transpose.reverse.map{|x|puts x*""}  Try it online! a full program which assumes there will be no newlines in the string. # J, 22 bytes 2|.@|:@#[[;.1~1,2~:/\]  Try it online! # APL (Dyalog), 37 bytes {⊃,/⍵↑⍨¨-⌈/≢¨⍵}'(.)\1*'⎕S{2/⍪⍵.Match}  Try it online! # q/kdb+, 53 bytes Solution: {(|)(+)(,/)(max(#:)each c)$(+)2#(,)c:((&)differ x)_x}


Example:

 q){(|)(+)(,/)(max(#:)each c)$(+)2#(,)c:((&)differ x)_x}"BBPPPPxxGGGGKKKKKKKkkkkEEeeEEEeeEEEEEOOO8####xxXXX" " KK " " KK " " KK EE " " PP GGKKkk EE ## " " PP GGKKkk EE EEOO ## XX" "BBPPxxGGKKkkEEeeEEeeEEOO ##xxXX" "BBPPxxGGKKkkEEeeEEeeEEOO88##xxXX"  Explanation: {reverse flip raze (max count each c)$flip 2#enlist c:(where differ x)_x} / ungolfed function
{                                                                       } / lambda function
(where differ x)    / indices where x differs
_   / cut at these points aabbbc -> "aa","bbb","c"
c:                    / save in variable c
enlist                       / put this list in another list
2#                             / take two from this list (duplicate)
flip                                / rotate columns/rows
(max count each c)                                     / find the longest run of characters
$/ whitespace pad lists to this length raze / reduce down lists flip / rotate columns/rows reverse / invert so buildings are on the ground  # Perl 5, 92 + 1 (-p) = 93 bytes while(s/(.)\1*//){$a[$_].=($"x($i-length$a[$_])).$1x2for 1..length$&;$i+=2}say while$_=pop@a  Try it online! # JavaScript (V8), 98 bytes x=>x.replace(/(.)\1*/g,y=>s=s.replace(/$/mg,_=>y[i]+y[i--]||'  ',i=n),s=
.repeat(n=x.length))&&s


Try it online!

# JavaScript (V8), 100 bytes

x=>x.replace(/(.)\1*/g,(y,c)=>s=s.replace(/\$/mg,_=>y[i--]?c+c:'  ',i=n),s=
.repeat(n=x.length))&&s
`

Try it online!