Oreoorererereoo

Question

Oreoorererereoo

Given an input string that is similar to the word "oreo", give an ASCII representation of the cookie that is as wide as the input string (to ensure cookie stability).

Rules

• The input is lowercase, a non-empty string with no whitespace containing any combination of the strings "o" and "re", and containing only those strings.
• The string "o" represents the solid cookie, while the string "re" represents the filling.
• The output must be a stacked cookie that is as wide as the input string.
• The output may not be an array of strings
• The cookie must overlap the filling by one character on each side
• The characters used for the output don't have to match the output below (█ and ░), they just have to be different non-whitespace characters for the two parts of the cookie
• The whitespace padding on the left side of the filling is required, and any trailing whitespace is optional

Examples

Input: oreo
Output:
████
 ░░ 
████

Input: o
Output:
█

Input: re
Output: (two spaces)


Input: rere
Output:
 ░░ 
 ░░ 

Input: oreoorererereoo
Output:
███████████████
 ░░░░░░░░░░░░░ 
███████████████
███████████████
 ░░░░░░░░░░░░░ 
 ░░░░░░░░░░░░░ 
 ░░░░░░░░░░░░░ 
 ░░░░░░░░░░░░░ 
███████████████
███████████████

Since this is code golf the shortest answer wins, good luck :)

Answer 1

Jelly,  16 14  13 bytes

-1 Thanks to Erik the Outgolfer

OḂƇẒṁ€aØ.¦€⁶Y

Uses 1 for the cream and 0 for the cookie.

Try it online!

How?

OḂƇẒṁ€aØ.¦€⁶Y - Main Link: list of characters, V    e.g. 'orereo'
O             - ordinal (vectorises)                     [111,114,101,114,101,111]
  Ƈ           - filter keep those for which:
 Ḃ            -   modulo 2                               [111,    101,    101,111]
   Ẓ          - is prime? (vectorises)                   [  0,      1,      1,  0]
    ṁ€        - mould each like V                        [[0,0,0,0,0,0],[1,1,1,1,1,1],[1,1,1,1,1,1],[0,0,0,0,0,0]]
          €   - for each:
         ¦    -   sparse application...
       Ø.     -   ...to indices: literal [0,1] (0 is the rightmost index, 1 is the leftmost)
      a       -   ...apply: logical AND with:
           ⁶  -               space character           [[0,0,0,0,0,0],[' ',1,1,1,1,' '],[' ',1,1,1,1,' '],[0,0,0,0,0,0]]
            Y - join with newline characters            [0,0,0,0,0,0,'\n',' ',1,1,1,1,' ','\n',' ',1,1,1,1,' ','\n',0,0,0,0,0,0]
              - implicit print                       ...smashes everything together:
              -                                         000000
              -                                          1111 
              -                                          1111 
              -                                         000000

---

Previous 16 byter:

ḟ”eẋ€Ly@Ø.¦€⁾r Y

Uses r for the cream and o for the cookie.

Try it online!

Answer 2

Pepe, 364 bytes

Unfortunately the online interpreter does not take care of compressing comments, hence all o characters will be replaced by a space.. Neither the spaces nor the o are necessary, so this could be 295 bytes, but I like it more this way:

rEeEEeeEeEororEEoreoreeeEeeeeeorEEEEeoREeoreorEeEEeEEEEororEEoreorEEEEEoREeoreorEeEEEeeEeororEEoreoReoREoREEEeoREEEEEoreorEorEEEeorEEEEEoreEoREeoreoREEeoREEEEeEeeoREEEeoREeeEoREEEeoREEEEEEEorEEEeEorEEEeoREoREEEeoREEEEEoREEoReoreorEEEeEoREEEEEEeorEEEeoReEoREoREEEeoREEoReoroReEeoREoREEEeorEEEEeoReeoREEEeoREeeEoREEEeoREEEEEEEoreoReoReoREoREEEeoREEEEEoreeeeeEeEeoRee

Try it online!

Ungolfed

There might be some golfing oppurtunities with flags which I missed, but I'm done for now:

# "function" for 'e'
rEeEEeeEeE rrEE
  re          # remove duplicated argument
  reeeEeeeee  # print space
  rEEEEe      # decrement counter twice
REe re

# "function" for 'o'
rEeEEeEEEE rrEE
  re      # remove duplicated argument
  rEEEEE  # increment counter
REe re

# "function for 'r'
rEeEEEeeEe rrEE
  re Re              # remove duplicated argument & char
  RE REEEe REEEEE    # push 1
  re rE rEEEe rEEEEE # replace 1
  reE                # goto 1
REe re

# Main

REEe REEEEeEee                # read input & reverse
REEEe REeeE REEEe REEEEEEE    # push length-1 & move to r

rEEEeE rEEEe # dummy loop-var (fucking do-whiles...)
RE REEEe REEEEE REE  # while [label-1]

  # Call the right procedure depending on current character,
  # sets stacks up as follows:
  #   R [ .... *currentChar ]
  #   r [ (N-1) *count ]
  Re re          # pop 1 & loop-counter
  rEEEeE         # duplicate counter
  REEEEEEe rEEEe # copy current char to other stack
  ReE            # jeq to 'o'-label or 'e'-label

  # Output currentChar count times:
  RE REEEe REE # while [label-0]:
    Re         #   pop 0
    rReEe      #   print character
    RE REEEe   #   push 0
    rEEEEe     #   decrement counter
  Ree

  REEEe REeeE REEEe REEEEEEE  # push length-1 & move to r
  re Re Re                    # pop 0, counter and 9((((currentChar
  RE REEEe REEEEE             # push 1
  reeeeeEeEe                  # print new-line

Ree

Answer 3

Canvas, 19 18 17 bytes

e ∙╋
：r≠＊┤］；Ｌ×⁸↔⁸

Try it here!

Uses the annoyingly long code of ：r≠＊┤］ to remove rs from the input..

Answer 4

Japt -R, 16 15 bytes

re ¬£çX sX²èrÃû

Try it

                    :Implicit input of string U
re                  :Remove all "e"s
   ¬                :Split to array of characters
    £               :Map each X
     çX             :  Repeat X to the length of U
        s           :  Slice from index
         X²         :   Duplicate X
           èr       :   Count the occurrences of "r"
             Ã      :End map
              û     :Centre pad each element with spaces to the length of the longest
                    :Implicitly join with newlines and output

---

Alternatives

re ¬ËpUÊaD²èrÃû
re ¬£îX rr²i^Ãû

Answer 5

C# (Visual C# Interactive Compiler), 95 bytes

n=>n.Replace("o",new String('-',n.Length)+"\n").Replace("re"," ".PadRight(n.Length-1,'|')+"\n")

Try it online!

Alternative using Aggregate, 108 bytes

n=>n.Aggregate("",(d,c)=>d+(c<102?"":c<112?new String('-',n.Length)+"\n":" ".PadRight(n.Length-1,'|')+"\n"))

Try it online!

Answer 6

05AB1E, 18 17 16 bytes

'eKεD'rQ2*Igα×}.c

-1 byte thanks to @Emigna

Uses o for the cookie and r for the filling.

Try it online or verify all test cases.

Explanation:

'eK                 '# Remove all "e" from the (implicit) input
                     #  i.e. "orereo" → "orro"
   ε         }       # Map all characters to:
    D                #  Duplicate the current character
     'rQ            '#  Check if it's an "r" (1 if truthy; 0 if falsey)
                     #   i.e. "r" → 1
                     #   i.e. "o" → 0
        ·            #  Double that
                     #   i.e. 1 → 2
                     #   i.e. 0 → 0
         Ig          #  Take the length of the input
                     #   i.e. "orereo" → 6
           α         #  Take the absolute difference between the two
                     #   i.e. 2 and 6 → 4
                     #   i.e. 0 and 6 → 6
            ×        #  Repeat the duplicated character that many times
                     #   i.e. "r" and 4 → "rrrr"
                     #   i.e. "o" and 6 → "oooooo"
              .c     # Then centralize it, which also imlicitly joins by newlines
                     # (and the result is output implicitly)
                     #  i.e. ["oooooo","rrrr","rrrr","oooooo"]
                     #   → "oooooo\n rrrr\n rrrr\noooooo"

Answer 7

R, 106 bytes

function(s,N=nchar(s)){m=rep(el(strsplit(gsub('re',0,s),'')),e=N)
m[m<1&seq(m)%%N<2]=' '
write(m,1,N,,"")}

Try it online!

• -12 bytes thanks to @Giuseppe

---

Previous version with explanation :

R, 118 bytes

function(s,N=nchar(s)){m=t(replicate(N,el(strsplit(gsub('re',0,s),''))))
m[m<1&row(m)%in%c(1,N)]=' '
write(m,1,N,,'')}

Try it online!

• -1 byte thanks to @Giuseppe

Unrolled code and explanation :

function(s){                       # s is the input string, e.g. 'oreo'

  N = nchar(s)                     # store the length of s into N, e.g. 4

  s1 = gsub('re',0,s)              # replace 're' with '0' and store in s1, e.g. 'o0o'

  v = el(strsplit(s1,''))          # split s1 into a vector v of single characters
                                   # e.g. 'o','0','o'

  m = replicate(N,v)               # evaluate N times the vector v and arrange 
                                   # the result into a matrix m (nchar(s1) x N)
                                   # e.g. 
                                   # 'o' 'o' 'o' 'o' 
                                   # '0' '0' '0' '0' 
                                   # 'o' 'o' 'o' 'o' 


  m = t(m)                         # transpose the matrix

  m[m<1 & row(m)%in%c(1,N)] = ' '  # substitute the zeros (i.e. where < 1) 
                                   # on the 1st and last row of the matrix with ' ' (space)
                                   # e.g. 
                                   # 'o' ' ' 'o' 
                                   # 'o' '0' 'o' 
                                   # 'o' '0' 'o' 
                                   # 'o' ' ' 'o'

  write(m,1,N,,'')                 # write the matrix to stdout (write function transposes it)
                                   # e.g.
                                   # oooo
                                   #  00 
                                   # oooo
}

Answer 8

Retina, 74 73 bytes

I feel like I haven't posted an answer in a very long time. Well, here I am. Also, Retina has changed a lot, and I feel like I suck at it now.

.+
$0$.0
(\d+)
*
e

o|r
$&¶
_$

+(/_/&`o¶
oo¶
_$

)/_/&`r¶
rr¶
¶$

m`^r
 

Try it online!

Answer 9

C (gcc), 135 113 109 104 bytes

• Saved twenty-two twenty-seven bytes thanks to NieDzejkob.
• Saved four bytes thanks to ceilingcat.

#define $ putchar(33
O(char*r){for(char*e,*o=r,x;*r;$-23))for(x=*r++>111,e=x?$-1),r++,o+2:o;*e++;$+x));}

Try it online!

Answer 10

Retina, 21 bytes

r

L$`.
$.+*$&
\bee
 

Try it online! Explanation:

r

Delete the rs.

L$`.
$.+*$&

List each letter on its own line repeated to the length of the original input.

\bee
 

Replace the first two ees on each line with a space.

Answer 11

JavaScript ES6, 103 bytes

Using replace 103 bytes:

x=>x.replace(/o/g,"-".repeat(s=x.length)+`
`).replace(/re/g," "+"|".repeat(s>1?s-2:0)+` 
`).slice(0,-1)

Try it online!

Using split and map 116 bytes:

x=>x.split("re").map(y=>("-"[h='repeat'](r=x.length)+`
`)[h](y.length)).join(" "+"|"[h](r>1?r-2:0)+` 
`).slice(0,-1)

Try it online!

Answer 12

Python 3, 77 bytes

lambda x:x.replace("o","-"*len(x)+"\n").replace("re"," "+'.'*(len(x)-2)+"\n")

Try it online!

Answer 13

Java 11, 110 bytes

s->{int l=s.length();return s.replace("re"," "+"~".repeat(l-(l<2?1:2))+"\n").replace("o","=".repeat(l)+"\n");}

Uses = for the cookie and ~ for the filling.

Try it online.

Explanation:

s->{                       // Method with String as both parameter and return-type
  int l=s.length();        //  Get the length of the input
  return s                 //  Return the input
          .replace("re",   //  After we've replaced all "re" with:
            " "            //   A space
            +"~".repeat(l-(l<2?1:2))
                           //   Appended with length-2 amount of "~"
                           //   (or length-1 if the input-length was 1)
            +"\n")         //   Appended with a newline
          .replace("o",    //  And we've also replaced all "o" with:
            "=".repeat(l)  //   Length amount of "="
            +"\n");}       //   Appended with a newline

---

The above solution uses a replace. The following maps over the characters of the input instead:

Java 11, 113 112 bytes

s->s.chars().forEach(c->{if(c>101)System.out.println((c>111?" ":"")+(""+(char)c).repeat(s.length()-2*(~c&1)));})

-1 byte thanks to @Neil.

Try it online.

Explanation:

s->                           // Method with String parameter and no return-type
  s.chars().forEach(c->{      //  Loop over the characters as codepoint-integers
    if(c>101)                 //   If it's not an 'e':
      System.out.println(     //    Print with trailing newline:
       (c>111?                //     If it's an 'r'
         " "                  //      Start with a space
        :                     //     Else (it's an 'o' instead)
         "")                  //      Start with an empty string
       +(""+(char)c).repeat(  //     And append the character itself
          .repeat(            //     Repeated the following amount of times:
           s.length()         //      The input-length
           -2*(~c&1)));})     //      Minus 2 if it's an "r", or 0 if it's an "o"

Answer 14

JavaScript, 72 65 64 bytes

s=>s.replace(/.e?/g,([x,y])=>(y?`
 `:`
`).padEnd(s.length+!y,x))

Try it online

Answer 15

Perl 5 -p, 47 bytes

s|o|X x($i=y///c).$/|ge;s|re|$".O x($i-2).$/|ge

Try it online!

Answer 16

Powershell, 71 69 66 bytes

-2 bytes thanks @Veskah

-3 bytes thanks @AdmBorkBork

$l=$args|% le*
switch($args|% t*y){'o'{'#'*$l}'r'{" "+'%'*($l-2)}}

Less golfed test script:

$f = {

$l=$args|% length
switch($args|% t*y){
    'o'{'#'*$l}
    'r'{" "+'%'*($l-2)}
}

}

@(

,(
'oreo',
'####',
' %%',
'####'
)
,(
'o',
'#'
)
,(
're',
' '
)
,(
'rere',
' %%',
' %%'
)
,(
'oreoorererereoo',
'###############',
' %%%%%%%%%%%%%',
'###############',
'###############',
' %%%%%%%%%%%%%',
' %%%%%%%%%%%%%',
' %%%%%%%%%%%%%',
' %%%%%%%%%%%%%',
'###############',
'###############'
)

) | % {
    $s,$expected = $_
    $result = &$f $s
    "$result"-eq"$expected"
    # $result # uncomment this line to display a result
}

Output:

True
True
True
True
True

Answer 17

PHP, 100 99 93 bytes

$l=strlen($i=$argv[1]);$r=str_repeat;echo strtr($i,[o=>$r(X,$l)."
",re=>' '.$r(o,$l-2)."
"]);

Try it online!

OUCH. PHP's waaaay_too_long function names strike again!

Output:

$php oreo.php oreo
XXXX
 oo
XXXX

$php oreo.php o
X

$php oreo.php rere
 oo
 oo

$ php oreo.php oreoorererereoo
XXXXXXXXXXXXXXX
 ooooooooooooo
XXXXXXXXXXXXXXX
XXXXXXXXXXXXXXX
 ooooooooooooo
 ooooooooooooo
 ooooooooooooo
 ooooooooooooo
XXXXXXXXXXXXXXX
XXXXXXXXXXXXXXX

Answer 18

Mathematica, 111 91 bytes

#~StringReplace~{"o"->"O"~Table~(n=StringLength@#)<>"\n","re"->" "<>Table["R",n-2]<>" \n"}&

Try It Online!

This was majorly shortened thanks to Misha's edits.

---

My original code:

(z=StringRepeat;n=StringLength@#;#~StringReplace~{"o"->"O"~z~n<>"\n","re"->" "<>If[n>2,z["R",n-2],""]<>" \n"})&

This code is not very fancy but it seems too expensive to convert away from strings and then back or to do anything else clever.

In particular, with only 3-4 commands that have the name String, my original approach couldn't save bytes at all by trying to abstract that away. For example, the following is 129 bytes:

(w=Symbol["String"<>#]&;z=w@"Repeat";n=w["Length"]@#;#~w@"Replace"~{"o"->"O"~z~n<>"\n","re"->" "<>If[n>2,z["R",n-2],""]<>" \n"})&

Answer 19

PHP, 96 87 85 bytes

Thanks to @gwaugh -9 Bytes
Thanks to @manatwork -2 Bytes

<?=strtr($i=$argv[1],[o=>($r=str_repeat)(X,$l=strlen($i))."
",re=>" {$r(o,$l-2)}
"]);

Try it online!

Try it online! (87 Bytes)

Try it online (original 97 bytes submition)!

---

And a recursive function

PHP, 135 bytes

function f($w,$x=0){$f=str_repeat;echo($x<($l=strlen($w)))?($w[$x]=='o')?$f(█,$l)."
".f($w,$x+1):" ".$f(░,$l-2)."
".f($w,$x+2):"";}

Try it online! (recursive)

Answer 20

Perl 6, 37 bytes

{m:g/o|r/>>.&({S/rr/ /.say}o*x.comb)}

Try it online!

Anonymous code block that takes a string and prints the oreo, with o as the cookie and r as the cream.

Explanation:

{                                   }   # Anonymous code block
 m:g/o|r/                               # Select all o s and r s
         >>.&(                     )    # Map each letter to
                            *x.comb     # The letter padded to the width
               S/rr/ /                  # Substitute a leading rr with a space
                      .say              # And print with a newline

Answer 21

Pyth, 28 bytes

FNzIqN"o"*lzN)IqN"r"+d*-lz2N
FNz                              For each value, N, in input
   IqN"o"                        if the character is "o"
         *lzN                    return the character times the length of the input
             )                   end if
              IqN"r"             if the character is "r"
FNzIqN"o"*lzN)IqN"r"+d*-lz2N
                        *-lz2N   return the character times length - 2
                    +d           padded on the left with " "

Try it here! This one uses a loop.

Pyth, 30 bytes

(As string replace)

::z"o"+*lz"="b"re"++d*-lz2"~"b
 :z"o"                           With the input, replace "o" with
       *lz"="                    "=" times the length of the input
      +      b                   and a newline added to the end
:             "re"               With the input, replace "re" with
                     *    "~"    "~" times
                      -lz2       the length of the input minus 2
                   +d            padded on the left with " "
                  +          b   and a newline added to the end

Try it here! This one uses string replacement.

I really like python (it's what I wrote my original test scripts in), so I thought I'd do a pyth entry for fun :)

Answer 22

Charcoal, 19 bytes

Ｆθ≡ιo⟦⭆θ#⟧e«→Ｐ⁻Ｌθ²↙

Try it online! Link is to verbose version of code. Explanation:

Ｆθ

Loop through the characters of the input string.

≡ι

Switch on each character.

o⟦⭆θ#⟧

If it's an o then print the input string replaced with #s on its own line.

e«→Ｐ⁻Ｌθ²↙

If it's an e then move right, print a line of -s that's two less than the length of the input string, then move down and left.

Answer 23

C# (Visual C# Interactive Compiler), 71 bytes

s=>s.Aggregate("",(a,c)=>a+(c>111?" ":"\n".PadLeft(s.Length+c/5-21,c)))

Try it online!

Borrowed some ideas from on Embodiment of Ignorance's answer for sure.

-6 bytes thanks to @ASCIIOnly!

The overall concept is to compute a string aggregate over the input characters following these rules:

• If an r is encountered, append a single space character for indentation. We know the next character will be an e.
• If an o or an e is encountered, generate a string by repeating the current character a specific number of times and prepending it to a newline or some padding and a newline.
• The number of times to repeat is determined by length of input string and whether the current line is indented.
• The PadLeft function is used to generate the repeating character string.

The result is the concatenation of all of these strings.

Answer 24

Bash, 87 bytes

Without sed:

f(){ printf %$1s|tr \  $2;}
c=${1//o/`f ${#1} B`
}
echo "${c//re/ `f $[${#1}-2] F` 
}"

Thanks to @manatwork.

With sed (90 bytes):

f(){ printf %$1s|tr \  $2;}
echo $1|sed "s/o/`f ${#1} B`\n/g;s/re/ `f $[${#1}-2] F` \n/g"

Answer 25

Ruby, 62 60 bytes

->s{s.gsub /./,?r=>" #{(?**z=s.size)[0..-3]}
",?o=>?O*z+?\n}

Try it online!

Uses O for the cookie, * for the filling.

-1 thanks to @manatwork pointing out a silly mistake and another -1 due to relaxation of the rules about whitespaces.

Answer 26

Dart, 120 106 107 bytes

f(s)=>s.replaceAll('o',''.padRight(s.length,'#')+'\n').replaceAll('re',' '.padRight(s.length-1,'-')+' \n');

Try it online!

• +1 byte : Added trailing whitespace

Answer 27

Python 2, 77 76 72 bytes

lambda i:'\n'.join((x*len(i),' '+x*(len(i)-2))[x>'o']for x in i if'e'<x)

Try it online!

The outer part of the cookie is 'o' and the filling is 'r'.

Answer 28

C# (.NET Core), 143 bytes

Without LINQ.

p=>{var q="";foreach(char c in p){if(c!='e'){for(var j=0;j<p.Length;j++)q+=(j<1|j>p.Length-2)&c>'q'?" ":c<'p'?"█":"░";q+="\n";}}return q;};

Try it online!

Answer 29

Clojure, 137 bytes

(fn[f](let[w(count f)r #(apply str(repeat % %2))](clojure.string/join"\n"(replace{\o(r w \#)\e(str \ (r(- w 2)\-) \ )}(remove #{\r}f)))))

I'm not using the nice characters in the printout in the golfed version since those are expensive. Returns a string to be printed.

Try it online!

See below for explanation.

Pre-golfed:

; Backslashes indicate a character literal
(defn oreo [format-str]
  (let [width (count format-str)

        ; A helper function since Clojure doesn't have built-in string multiplication
        str-repeat #(apply str (repeat % %2))

        ; Define the layers
        cookie (str-repeat width \█)
        cream (str \ (str-repeat (- width 2) \░) \ )]

    (->> format-str ; Take the input string,
         (remove #{\r}) ; remove r for simplcity,
         (replace {\o cookie, \e cream}) ; replace the remaining letters with the layers,
         (clojure.string/join "\n")))) ; and join the layers together with newlines

Answer 30

x86-64 machine code (Linux), 97 bytes

0000000000000000 <oreo_asm>:
   0:   56                      push   %rsi
   1:   57                      push   %rdi

0000000000000002 <len>:
   2:   48 ff c7                inc    %rdi
   5:   80 3f 00                cmpb   $0x0,(%rdi)
   8:   75 f8                   jne    2 <len>
   a:   49 89 fc                mov    %rdi,%r12
   d:   5f                      pop    %rdi
   e:   49 29 fc                sub    %rdi,%r12
  11:   4d 31 f6                xor    %r14,%r14
  14:   eb 18                   jmp    2e <outer_loop.skip>

0000000000000016 <extra>:
  16:   41 c6 01 20             movb   $0x20,(%r9)
  1a:   c6 03 20                movb   $0x20,(%rbx)
  1d:   49 ff ce                dec    %r14
  20:   eb 06                   jmp    28 <outer_loop>

0000000000000022 <newline>:
  22:   c6 06 0a                movb   $0xa,(%rsi)
  25:   48 ff c6                inc    %rsi

0000000000000028 <outer_loop>:
  28:   49 ff c6                inc    %r14
  2b:   48 ff c7                inc    %rdi

000000000000002e <outer_loop.skip>:
  2e:   44 8a 07                mov    (%rdi),%r8b
  31:   41 80 f8 65             cmp    $0x65,%r8b
  35:   74 df                   je     16 <extra>
  37:   45 84 c0                test   %r8b,%r8b
  3a:   74 23                   je     5f <done>
  3c:   48 89 f3                mov    %rsi,%rbx

000000000000003f <inner_loop>:
  3f:   44 88 06                mov    %r8b,(%rsi)
  42:   49 89 f1                mov    %rsi,%r9
  45:   48 ff c6                inc    %rsi
  48:   48 31 d2                xor    %rdx,%rdx
  4b:   48 89 f0                mov    %rsi,%rax
  4e:   48 2b 04 24             sub    (%rsp),%rax
  52:   4c 29 f0                sub    %r14,%rax
  55:   49 f7 f4                div    %r12
  58:   48 85 d2                test   %rdx,%rdx
  5b:   74 c5                   je     22 <newline>
  5d:   eb e0                   jmp    3f <inner_loop>

000000000000005f <done>:
  5f:   5e                      pop    %rsi
  60:   c3                      retq

This x86-64 function takes in the pointer to the input string in rsi and builds the output starting at the pointer in rdi (these are the registers used to pass the first two arguments from a C function on Linux). For convenience, I've written a C++ wrapper for this which also does nice input sanitization and prints the output. That code can be located here. This also shows the original nasm syntax assembly I wrote for this function (as well as the non-golfed version I got working first).

A few things to note is that this code doesn't respect any callee saved registers, which means that the C++ code likely will crash if run for a while after calling this function. On my machine it doesn't, but that's rather surprising. I also don't add a null byte to delimit the output string, and instead the space allocated for the output string is pre-filled with bytes. (If this isn't allowed I can add the null terminator at a cost of 3 bytes).

The logic for this code is essentially counting the length of the string, then building a line of this length for each 'o' and 'r' characters seen in the input string, and then for any 'e' character seen, replacing the first and last characters on the previous line with space characters.

I can't find anywhere online to compile and run a mix of C++ and nasm source code, so I might write some small wrapper code for this to prove it works. Otherwise you should be able to compile and run this with the makefile in the link I gave with the command:

$ make oreo ASM_FILE=oreo_golf.nasm
$ ./oreo oreoorererereoo --use_asm

I was able to format the assembly to something acceptable by gcc, so try it online!

Answer 31

Sed, 89 characters

(86 characters code + 3 characters command line option)

h;s/./E/g;H;g;s/\n//;:b;s/(o|re)([ore]+)(E*)/\1\3\n\2\3/;tb;s/o//g;s/reE(E*)E/ \L\1 /g

Commented with example for input “oreo”:

h        # copy to hold space
s/./E/g  # change into cookie: EEEE
H        # append to hold space
g        # copy back to pattern space: oreo␤EEEE
s/\n//   # remove the newline: oreoEEEE

:b                                 # split the os and res in separate lines
  s/(o|re)([ore]+)(E*)/\1\3\n\2\3/ # repeating the EEEE for each:
tb                                 # oEEEE␤reEEEE␤oEEEE

s/o//g                # remove the os: EEEE␤reEEEE␤EEEE
s/reE(E*)E/ \L\1 /g   # remove the res, cut edge, change fill: EEEE␤ ee ␤EEEE

Sample run:

bash-4.4$ sed -r 'h;s/./E/g;H;g;s/\n//;:b;s/(o|re)([ore]+)(E*)/\1\3\n\2\3/;tb;s/o//g;s/reE(E*)E/ \L\1 /g' <<< oreoorererereoo
EEEEEEEEEEEEEEE
 eeeeeeeeeeeee 
EEEEEEEEEEEEEEE
EEEEEEEEEEEEEEE
 eeeeeeeeeeeee 
 eeeeeeeeeeeee 
 eeeeeeeeeeeee 
 eeeeeeeeeeeee 
EEEEEEEEEEEEEEE
EEEEEEEEEEEEEEE

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Answer 32

C (gcc), -D$=putchar(33 + 88 = 102 bytes

O(char*r){for(char*e,*o=r,x;*r;$-x),$-23))for(x=*r++>111,e=x?$-1),r++,o+1:o;*++e;$+x));}

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Answer 33

APL (Dyalog Extended), 22 bytes^SBCS

Anonymous tacit prefix function. Cookies are o, cream is r.

'^rr'⎕R' '⍤1≢⊢⍤/⍪~'e'⍨

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…⍨ swap the input for:

 'e' this letter

…~ remove all the "e"s from:

 ⍪ the columnised argument

…⊢⍤/ replicate that single column to be as many columns as:

 ≢ the tally of characters in the input

…⍤1 on each row (lit. on each 1D sub-array):

 '^rr'⎕R' ' PCRE Replace leading "rr" with a single space

Answer 34

Brainfuck, 276 268 bytes

Technicaly as the brainfuck only has 8 instructions, that means it needs only 3 bits per character instead of 8 (that also means we could encode brainfuck in base64 with exactly 2 instructions per character), but I'll play fair and say it's 268 bytes long.

I was afraid of using more cells to set up the values to be displayed, so it can probably be shortened, at least I used one to get a shorter solution.

It was a fun challenge, thanks!

Edit: now featuring the 3rd rule!

Edit: moving the line feed from the datapool to the display part of the code allowed me to get down to 268 bytes.

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

Here's a more readable version:

The datapool is made like so: [first character, placeholder value used to move the input character count, repeated character, last character]

>>+<,[//start of the loop
<++++++++++[>-----------<-]>-//"o"
[>-<,>>+<<[-]//if it's not a "o", skip one character and:

>>>>[>]//go to the data pool and search for the end
>++++[<++++++++>-]//put " " in memory
+>//put 1 in memory
>++++++++++[<+++++++++++>-]<+>//put a "o" in memory
>++++[<++++++++>-]<//put another " " in memory
[<]<<<//get back to the beginning of the data pool and get back to the program

]>[[-]//else (so if it's a "o"):

>>>[>]//go to the data pool and search for the end
>+++++++[<+++++>-]//put a "#" in memory
+>//put 1 in memory
>+++++++[<+++++>-]//put a "#" in memory
>+++++++[<+++++>-]<//put a "#" in memory
[<]<<//get back to the beginning of the data pool and get back to the program

]+<,>>+<<]//loop until the end of the input line and count the number of characters

>>//go to the number of characters variable
--//substract 2 to it because the first and last characters take one space each
[>>>+<<<-]//copy the value to the next place
>>[.>//start of the loop and display of the first character
-[>>>>+<<<<->.<]//display of the repeated character while copying the character count to the next line
>>.//display of the last character
[-]++++++++++.//display a line feed
>]//loop until the end of the data pool

it works on copy.sh/brainfuck with default settings (link)

Answer 35

SNOBOL4 (CSNOBOL4), 136 bytes

	S =INPUT
	N =SIZE(S)
S	S ('o' | 're') . X REM . S	:F(END)
	OUTPUT =IDENT('o',X) DUPL(X,N)	:S(S)
	OUTPUT =' ' DUPL(0,N - 2) ' '	:(S)
END

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Answer 36

Batch, 133 bytes

@set/ps=
@set o=%s:r=o%
@set o=%o:e=o%
@set r=%o:o=-%
@set r= %r:~2%
@set s=%s:e= %
@for %%c in (%s:o=o %)do @call echo %%%%c%%

Takes input on STDIN. Explanation:

@set/ps=

Input the string.

@set o=%s:r=o%
@set o=%o:e=o%

Make a copy of the string with all of the rs and es replaced with os.

@set r=%o:o=-%
@set r= %r:~2%

Make a copy of the string of os, change them to -s and replace the first two with a space.

@set s=%s:e= %
@for %%c in (%s:o=o %)do @call echo %%%%c%%

Change the es into spaces and also add a space after each o, allowing the os and rs to be iterated over, printing the appropriate string each time.

Answer 37

CJam, 27 bytes

q:T'e-{i2%T,'9*_:)2>S\+?N}/

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Explanation:

q                            read input:                    | "oreoo"
 :T                          store in variable T            | "oreoo", T="oreoo"
   'e-                       remove all 'e' characters:     | "oroo"
      {                  }/  for each character:            | 'o
       i                       get Unicode value:           | 79
        2%                     modulo 2:                    | 1
          T                    push original input:         | 1 "oreoo"
           ,                   get length:                  | 1 5
            '9*                repeat '9' that many times:  | 1 "99999"
               _               duplicate:                   | 1 "99999" "99999"
                :)             increment every character:   | 1 "99999" ":::::"
                  2>           remove first two characters: | 1 "99999" ":::"
                    S\+        prepend a space:             | 1 "99999" " :::"
                       ?       conditional:                 | "99999"
                        N      add a newline:               | "99999" N
                             (implicit output)              | "99999" N " :::" N "99999" N "99999" N

Answer 38

Lua, 112 bytes

s=io.read()for t in s:gmatch"."do io.write(t=="o"and t:rep(#s).."\n"or(t=="r"and" "..t:rep(#s-2).." \n"or""))end

---

Explanation

s = io.read() -- Reads the input
for t in s:gmatch(".") do -- For each character in the input, call it 't' and enter the loop
 io.write(t=="o" and t:rep(#s).."\n" or (t=="r" and " "..t:rep(#s-2).." \n" or ""))

          t=="o" and  -- if 't' is equal to "o" then
                     t:rep(#s).."\n"  -- you have to write 't' repeated N times, where N is
                                      -- is the size of the input s, appended by a new line
                                     or (t=="r" and  -- else, if 't' is equal to "r" then
                                                    " "..t:rep(#s-2).." \n"  -- write a space, followed by 't' repeated N-2 times,
                                                                             -- where N is the size of the input.
                                                                             -- followed by a space and a new line.
                                                                            or "" -- else, just don't write anything.

end

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Answer 39

JavaScript, 70 bytes

x=>x.replace(/()()o|re/g,(_,o=1,s=' ')=>s.padEnd(x.length-o,+o)+s+`
`)

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Or... If we're allowed to use other characters for whitespace, we can get it down to 65 bytes:

x=>x.replace(/()()o|re/g,(_,o=1,s=' ')=>s.padEnd(x.length-o,+o)+s+`
`)

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...but I don't think that counts.

Answer 40

JavaScript (Node.js), 65 bytes

s=>s[R='replace'](/.e?/g,x=>s[R](/./g,x[0])[R](/^r|r$/g,' ')+`
`)

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Thanks to Yair Rand, saves 2 bytes.

Answer 41

Python 2, 70 75 bytes

lambda s:"\n".join((c*(len(s)-2*(c<'f'))).center(len(s))for c in s if'r'>c)

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Uses 'o' for cookie, 'e' for filling. Returns a string.

EDIT: Saved 4 bytes thanks to Jo King.

Explanation:

# anonymous function, returns a string seperated by line breaks
lambda s:"\n".join(.......................................................)
                                                        # iterate through string, return each non-'r' character
                                                        for c in s if'r'>c
                    # duplicate the character until it is the length of the original string,
                    # minus two characters if the current letter is 'e'
                    c*(len(s)-2*(c<'f'))
                   # pad the string with spaces out to the length of the original string
                   (....................).center(len(s))

Answer 42

Python 2, 64 bytes

s=input()
for c in s:
	if'n'<c:j='o'<c;print' '*j+c*(len(s)-2*j)

A full program which prints the result.

Try it online! Or see the test-suite.

Answer 43

Tcl, 216 bytes

set s [gets stdin]
set l [string map {o { o } re { re }} $s]
set n [string length $s]
foreach i $l {
if {$i eq "o"} {
puts [string repeat █ $n]
} elseif {$i eq "re"} {
puts " [string repeat ░ [expr {$n-2}]] "
}
}

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Answer 44

Stax, 13 bytes

äNëY┬2-]‼Cv6╤

Run and debug it

Uses "o" and "r" characters for the layers.

Answer 45

Excel, 93 bytes

=SUBSTITUTE(SUBSTITUTE(A4,"o",REPT("O",LEN(A4))&"
"),"re"," "&REPT("r",MAX(0,LEN(A4)-2))&"
")

MAX(0, is needed to handle o. Uses O and r.