Fit a word into an alphabet grid

Inspired by a meme I saw earlier today.

Challenge description

Consider an infinite alphabet grid:

ABCDEFGHIJKLMNOPQRSTUVWXYZ
ABCDEFGHIJKLMNOPQRSTUVWXYZ
ABCDEFGHIJKLMNOPQRSTUVWXYZ
ABCDEFGHIJKLMNOPQRSTUVWXYZ
ABCDEFGHIJKLMNOPQRSTUVWXYZ
...


Take a word (CODEGOLF in this example) and make it a subsequence of the grid, replacing unused letters by a space and removing letters at the end of the infinite grid altogether:

  C           O
DE G       O
L
F


Examples

STACKEXCHANGE

                  ST
A C       K
E                  X
C    H
A            N
G
E


ZYXWVUTSRQPONMLKJIHGFEDCBA

                         Z
Y
X
W
V
U
T
S
R
Q
P
O
N
M
L
K
J
I
H
G
F
E
D
C
B
A


F

     F


ANTIDISESTABLISHMENTARIANISM

A            N     T
I
D    I         S
E             ST
AB         L
I         S
H    M
E        N     T
A                R
I
A            N
I         S
M


Notes

• Trailing whitespaces are allowed.
• You don't need to pad the last any line with spaces. For example, if the input is ABC, you may output just ABC without 23 trailing spaces.
• You may assume input will match [A-Z]+ regex.
• Alternatively, you may use lower-case alphabet, in which case output will match [a-z]+.
• You must use a newline (\n, \r\n or equivalent) to separate lines, that is a list of strings is not a proper output format.
• This is a challenge, so make your code as short as possible!
• Are leading newlines allowed? Sep 2, 2017 at 13:46
• @EriktheOutgolfer Sure, as long as it doesn't mess up grid structure. Sep 2, 2017 at 13:49
• Would it be okay if a non-fatal error stops the program? Sep 2, 2017 at 21:07
• @Zacharý Although I can see how that could save some bytes, I think it's ugly and produces undesired, superfluous output. So no. EDIT: Unless you can make your program non-fatally exit through an exit code or something that wouldn't print exception stack trace or something similar to stderr. Sep 2, 2017 at 21:09
• Suggested test case: BALLOON (two adjacent characters that are the same). Sep 4, 2017 at 12:37

Husk, 15 bytes

TṪS?' €…"AZ"ġ>


Try it online!

Explanation

TṪS?' €…"AZ"ġ>  Implicit input, e.g. "HELLO"
ġ>  Split into strictly increasing substrings: x = ["H","EL","LO"]
…"AZ"    The uppercase alphabet (technically, the string "AZ" rangified).
Ṫ               Outer product of the alphabet and x
S?' €         using this function:
Arguments: character, say c = 'L', and string, say s = "EL".
€           1-based index of c in s, or 0 if not found: 2
S?'             If this is truthy, then c, else a space: 'L'
This gives, for each letter c of the alphabet,
a string of the same length as x,
containing c for those substrings that contain c,
and a space for others.
T                Transpose, implicitly print separated by newlines.


Java 10, 161159 152 bytes

s->{var x="";int p=0;for(var c:s)x+=p<(p=c)?c:";"+c;for(var y:x.split(";"))System.out.println("ABCDEFGHIJKLMNOPQRSTUVWXYZ".replaceAll("[^"+y+"]"," "));}


-2 bytes thanks to @Nevay.
-7 byte printing directly instead of returning a String, and converting to Java 10.

Explanation:"

Try it here.

s->{                      // Method with String parameter and no return-type
var x="";               //  Temp-String
int p=0;                //  Previous character (as integer), starting at 0
for(var c:s)            //  Loop (1) over the characters of the input
x+=p<(p=c)?           //   If the current character is later in the alphabet
//   (replace previous p with current c afterwards)
c                 //    Append the current character to Temp-String x
:                  //   Else:
";"+c;            //    Append a delimiter ";" + this character to Temp-String x
for(var y:x.split(";")) //  Loop (2) over the String-parts
System.out.println(   //   Print, with trailing new-line:
"ABCDEFGHIJKLMNOPQRSTUVWXYZ"
//    Take the alphabet,
.replaceAll("[^"+y+"]"," "));}
//    and replace all letters not in the String-part with a space


The first part of the method splits the input-word into parts with a delimiter.
For example: CODEGOLFCO;DEGO;L;F or BALLOONB;AL;LO;O;N.

The second part loops over these parts, and uses the regex [^...] to replace everything that isn't matched with a space.
For example .replaceAll("[^CO]"," ") leaves the C, and O, and replaces everything else with a space.

• Wouldn't it be B;AL;LO;O;N? Sep 4, 2017 at 12:59
• -2 bytes: for(char c:s)x+=p<(p=c)?c:";"+c;. Sep 4, 2017 at 13:24

C (gcc), 69 bytes

i;f(char*s){for(i=64;*s;putchar(*s^i?32:*s++))i+=i^90?1:puts("")-26;}


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Perl 5, 44 bytes

40 bytes code + 4 for -lF.

print map/$F[0]/?shift@F:$",A..Z while@F


Try it online!

• The link you've posted is for 46 bytes version.
– user72349
Sep 2, 2017 at 14:03
• @ThePirateBay Thanks!! I knew I hadn't updated something! Sep 2, 2017 at 14:06

Python 2, 92 bytes

f=lambda x,y=65:x and(y<=ord(x[0])and" "*(ord(x[0])-y)+x[0]+f(x[1:],-~ord(x[0]))or"\n"+f(x))


Try it online!

• 90 bytes
– ovs
Sep 2, 2017 at 21:20

JavaScript (ES6), 79

Edit As a leading newline is accepted, I can save 2 bytes

s=>eval("for(o='',v=i=0;c=s.charCodeAt(i);v%=27)o+=v++?c-63-v?' ':s[i++]:\n")


For 1 byte more, I can accept lowercase or uppercase input:

s=>eval("for(o='',v=i=0;c=s[i];v%=27)o+=v++?parseInt(c,36)-8-v?' ':s[i++]:\n")


Less golfed

s=>{
var i,o,c,v
for(o = '', v = 1, i = 0; c = s.charCodeAt(i); v %= 27)
o += v++ ? c-63-v ? ' ' : s[i++] : '\n'
return o
}


Test

f=s=>eval("for(o='',v=i=0;c=s.charCodeAt(i);v%=27)o+=v++?c-63-v?' ':s[i++]:\n")

function update() {
var i=I.value
i=i.replace(/[^A-Z]/gi,'').toUpperCase()
O.textContent=f(i)
}

update()
<input id=I value='BALLOON' oninput='update()' >
<pre id=O></pre>

• You can replace the \n with a literal newline inside backticks for -1 byte. Sep 2, 2017 at 20:41
• @JustinMariner no I can't, not inside the double quote in eval Sep 2, 2017 at 20:48
• Oh right, that's a shame. My bad. Sep 2, 2017 at 20:49

MATL, 24 23 bytes

''jt8+t1)wdh26X\Ys(26e!


Uses lowercase letters.

Try it at MATL Online!

Explanation

''     % Push empty string
jt     % Push input string. Duplicate
8+     % Add 8 to each char (ASCII code). This transforms 'a' 105,
% 'b' into 106, which modulo 26 correspond to 1, 2 etc
t1)    % Duplicate. Get first entry
wd     % Swap. COnsecutive differences.
h      % Concatenate horizontally
26X\   % 1-based modulo 26. This gives a result from 1 to 26
Ys     % Cumulative sum
(      % Write values (converted into chars) at specified positions
% of the initially empty string
26e    % Reshape into a 26-row char matrix, padding with char 0
!      % Transpose. Implicitly display. Char 0 is shown as space


Japt, 18 16 bytes

-2 bytes thanks to @Shaggy

;ò¨ £B®kX ?S:Z
·


Uppercase input only.

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Explanation

;


Switch to alternate variables, where B is the uppercase alphabet.

ò¨


Split the input string between characters where the first is greater than or equal to (¨) the second.

£


Map each partition by the function, where X is the current partition.

B®


Map each character in the uppercase alphabet to the following, with Z being the current letter.

kX


Remove all letters in the current partition from the current letter. If the current letter is contained in the current partition, this results in an empty string.

?S:Z


If that is truthy (not an empty string), return a space (S), otherwise return the current letter.

·


Join the result of the previous line with newlines and print the result.

• 10 bytes for r"[^{Z}]"S seems a bit ridiculous, but I can't find any better way either... Sep 2, 2017 at 20:34
• 17 bytes Sep 3, 2017 at 11:54
• 16 bytes Sep 3, 2017 at 19:44
• @Shaggy Good thinking with kX! Sep 3, 2017 at 20:01
• Actually I think you can change kX ?S:Z to oX ªS to save two bytes Sep 5, 2017 at 3:10

Pyth, 18 bytes

#pbVGp?JqhzNNd=>zJ


Try it here.

Leading newline in output, lowercase alphabet.

<2\¬0;œṗfȯ⁶$¥€@€ØAY  Try it online! • OI<1®; -> >2\0; to save one byte (I actually did >2\0;œṗµØAf€ȯ€⁶µ€Y for 18 too, which I personally find easier to parse) Sep 3, 2017 at 0:32 • @JonathanAllan I think that would fail for BALLOON or something. Sep 3, 2017 at 12:07 • You are correct, yes - so it would required another byte with something like <2\1;¬; oh well. Sep 3, 2017 at 13:47 • @JonathanAllan Anyways, I'll implement your idea in my answer...done. Sep 3, 2017 at 13:53 C (gcc), 91 63 bytes -28 thanks to ASCII-only _;f(char*s){for(_=64;*s;)putchar(++_>90?_=64,10:*s^_?32:*s++);}  Try it online! Previous: i,j;f(char*s){while(s[i]){for(j=65;j<91;j++)s[i]==j?putchar(s[i++]):printf(" ");puts("");}}  Yes, there's a shorter solution, but I noticed after I wrote this one... Try it online! • 82 bytes, 80 if leading newline is allowed Sep 5, 2017 at 22:24 • 73 Sep 5, 2017 at 22:30 • 63 Sep 5, 2017 at 23:08 Mathematica, 101 bytes StringRiffle[ Alphabet[]/.#->" "&/@ (Except[#|##,_String]&@@@ Split[Characters@#,#==1&@*Order])," ",""]&  Split the input into strictly increasing letter sequences, comparing adjacent letters with Order. If Order[x,y] == 1, then x precedes y in the alphabet and thus can appear on the same line. For each sequence of letters, create a pattern to match strings Except for those letters; #|## is a shorthand for Alternatives. Replace letters of the Alphabet that match the pattern with spaces. Illustration of the intermediate steps: "codegolf"; Split[Characters@#,#==1&@*Order] &@% Except[#|##,_String]&@@@ #&@% Alphabet[]/.#->" "&/@ %  {{"c", "o"}, {"d", "e", "g", "o"}, {"l"}, {"f"}} {Except["c" | "c" | "o", _String], Except["d" | "d" | "e" | "g" | "o", _String], Except["l" | "l", _String], Except["f" | "f", _String]} {{" "," ","c"," "," "," "," "," "," "," "," "," "," "," ","o"," "," "," "," "," "," "," "," "," "," "," "}, {" "," "," ","d","e"," ","g"," "," "," "," "," "," "," ","o"," "," "," "," "," "," "," "," "," "," "," "}, {" "," "," "," "," "," "," "," "," "," "," ","l"," "," "," "," "," "," "," "," "," "," "," "," "," "," "}, {" "," "," "," "," ","f"," "," "," "," "," "," "," "," "," "," "," "," "," "," "," "," "," "," "," "," "}}  Golfscript, 22 21 bytes Try it online! -1 byte thanks to careful final redefining of the n built-in. {.n>{}{' '\}if:n}%:n;  Explanation (with a slightly different version): {.n>{}{"\n"\}if:n}%:n; # Full program { }% # Go through every character in the string .n> if # If ASCII code is greater than previous... # (n means newline by default, so 1st char guaranteed to fit) {} # Do nothing {"\n"\} # Else, put newline before character :n # Redefine n as the last used character :n; # The stack contents are printed at end of execution # Literally followed by the variable n, usually newline # So because n is by now an ASCII code... # ...redefine n as the new string, and empty the stack  Retina, 80 bytes ^ ;¶ {;.* ¶;ABCDEFGHIJKLMNOPQRSTUVWXYZ ¶¶ ¶ )+;(.*)(.)(.*¶)\2$.1$*$2;$3 ;.*  Try it online! There is always exactly one leading newline. The code somewhat clunkily prepends the word with the alphabet along with a marker (semicolon). It then moves the marker up to the first letter of the word, while changing all other letters it passes into spaces. It also removes the first letter of the word. It repeats this until the first letter of the word isn't after the marker anymore. Then it clears that marker and the rest of the alphabet, and replaces it with a new line and the alphabet with a marker again. It keeps repeating this until the input word is empty, then it cleans up the last alphabet and marker, leaving the desired output. 05AB1E, 18 bytes ćIgµ¶?AvDyÊið?ë¼?ć  Try it online! Got trouble with 05AB1E ć (extract 1) leaving an empty string/list on the stack after the last element is extracted. This solution would be 1-2 bytes shorter if it weren't for that. ćIgµ¶?AvDyÊið?ë¼?ć Implicit input ć Extract the 1st char from the string Igµ While counter != length of the string ¶? Print a newline Av For each letter of the lowercased alphabet DyÊ Is the examined character different from the current letter? ið? If true, then print a space ë¼?ć Else increment the counter, print the letter and push the next character of the string on the stack  • Actually, ð, means "print a space and a newline". Sep 4, 2017 at 13:56 • You're right. Fixed the code to actually print a newline. Sep 4, 2017 at 14:00 Retina, 130 126 bytes $
¶A
{-2=
$' }TRL_o.$
+(?<=(.)*)((.).*¶(?<-1>.)*(?(1)(?!)).+\3.*$)$2
(?<=(.)*)((.).*¶(?<-1>.)*(?<-1>\3.*$)) ¶$2
}¶.*$ Try it online! Edit: Saved 4 bytes by using @MartinEnder's alphabet generator. Explanation: $
¶A
{-2=
$' }TRL_o.$


Append the alphabet.

+(?<=(.)*)((.).*¶(?<-1>.)*(?(1)(?!)).+\3.*$)$2


Align as many letters as possible with their position in the alphabet.

(?<=(.)*)((.).*¶(?<-1>.)*(?<-1>\3.*$)) ¶$2


Start a new line before the first letter that could not be aligned.

}¶.* Delete the alphabet, but then do everything over again until there are no misaligned letters. • This seems to print only one line, not aligning letters on subsequent lines. Sep 2, 2017 at 18:23 • @JustinMariner My bad, I made a typo in my last golf and failed to check it properly. – Neil Sep 2, 2017 at 19:29 q/kdb+, 48 45 bytes Solution: -1{@[26#" ";.Q.A?x;:;x]}@/:(0,(&)(<=':)x)_x:;  Try it online! Note: Link is to a K (oK) port of this solution as there is no TIO for q/kdb+. Examples: q)-1{@[26#" ";.Q.A?x;:;x]}@/:(0,(&)(<=':)x)_x:"STACKEXCHANGE"; ST A C K E X C H A N G E q)-1{@[26#" ";.Q.A?x;:;x]}@/:(0,(&)(<=':)x)_x:"BALLOON"; B A L L O O N  Explanation: Q is interpreted right-to-left. The solution is split into two parts. First split the string where the next character is less than or equal to the current: "STACKEXCHANGE" -> "ST","ACK","EX","CH","AN","G","E" Then take a string of 26 blanks, and apply the input to it at the indices where the input appears in the alphabet, and print to stdout. "__________________________" -> __________________ST______ Breakdown: -1{@[26#" ";.Q.A?x;:;x]}each(0,where (<=':)x) cut x:; / ungolfed solution -1 ; / print to stdout, swallow return value x: / store input as variable x cut / cut slices x at these indices ( ) / do this together (<=':)x / is current char less-or-equal (<=) than each previous (':)? where / indices where this is true 0, / prepended with 0 each / take each item and apply function to it { } / lambda function with x as implicit input @[ ; ; ; ] / apply[variable;indices;function;arguments] 26#" " / 26 take " " is " "... .Q.A?x / lookup x in the uppercase alphabet, returns indice(s) : / assignment x / the input to apply to these indices  Notes: • -3 bytes by replacing prev with the K4 version Powershell, 70 63 bytes -7 bytes thanks @Veskah args|%{if($_-le$p){$x;rv x}$x=("$x"|% *ht($_-65))+($p=$_)}
$x  Try it online! Explanation: For each character in the splatted argument: • Output string $x and clear $x value (rv is alias for Remove-Variable), if a code of the current character less or equivalent (-le) to a code of the previous character. • Append spaces and the current character to $x, store it to $x. Also it freshes a previous character value. Output last $x.

• 63 Bytes using splatting. Tried to use |% *ht to save some bytes but looks like it broke even. Oct 9, 2019 at 17:10

Hexagony, 91 bytes

A"$>}\$<>~,<.'~<\.';<.>{}\../<..>=<...-/.$/=*=32;.>~<>"){-\}.>_{A}\\../010<$..>;=~<$@{{Z'  More readably:  A "$ > } \
$< > ~ , < . ' ~ < \ . ' ; < . >$ ${ } \ . . / < . . > = < . . . - / .$ / = * = 3 2 ;
. > ~ < > " ) { - \
} . > _ { A } \ \
. . / 0 1 0 < $. . > ; = ~ <$ @ { { Z '


And with color paths:

Try it online!

Jelly, 24 21 bytes

3 bytes thanks to Erik the Outgolfer.

O64;I%26’⁶ẋÐ€;"⁸Ẏs26Y


Try it online!

• I believe this fails for input "BALLOON" - the repeated characters are on the same line. Sep 2, 2017 at 18:03

SOGL V0.12, 22 bytes

±E⁄Z*{@;eJι=?Xē}}¹∑z⁄n


Try it Here!

JavaScript (ES6), 87 bytes

f=([...s])=>s[0]?(g=i=>i>35?
+f(s):(i-parseInt(s[0],36)?" ":s.shift())+g(i+1))(10):""


Accepts uppercase or lowercase input. Output match the case of the input.

Tests

f=([...s])=>s[0]?(g=i=>i>35?
+f(s):(i-parseInt(s[0],36)?" ":s.shift())+g(i+1))(10):""

;O.innerText=["CODEGOLF","STACKEXCHANGE","F","ZYXWVUTSRQPONMLKJIHGFEDCBA","ANTIDISESTABLISHMENTARIANISM"]
.map(f).join("=".repeat(26)+"\n")
<pre id=O>

Husk, 222119 17 bytes

-2 bytes thanks to Zgarb (and a new language feature)

mȯΣẊṠ:ȯR' ←≠:'@ġ>


Try it online!

Explanation

               ġ>    Group into increasing sublists
mȯ                   To each sublist apply the following three functions
:'@      ¹Append '@' (The character before 'A') to the start
Ẋ                 ²Apply the following function to all adjacent pairs
≠            Take the difference of their codepoints
←             Minus 1
ȯR'               Repeat ' ' that many times
Ṡ:                  Append the second argument to the end.
Σ                  ³concatenate

• 19 bytes with the new binary overloading of Ṡ. Sep 2, 2017 at 14:21
• @Zgarb Wow fancy! Sep 2, 2017 at 14:23

q@(w:y)!(x:z)|w==x=x:y!z|1<2=min ' 'x:q!z
x!_=x
a=['A'..'Z']++'\n':a
(!a)


Saved 1 byte thanks to Laikoni!

Try it online.

1. The Hugs interpreter allows me to save one more byte by doing (!cycle$['A'..'Z']++"\n") instead of: (!cycle(['A'..'Z']++"\n")), but GHC does not like the former. (This is now obsolete; Laikoni already rewrote that line in a way that saved 1 byte.) 2. Apparently, Hugs also does not require parentheses around the list pattern matcher, so I could save two more bytes going from: q@(w:y)!(x:z) to q@(w:y)!x:z. • You can save a byte with a=['A'..'Z']++'\n':a;(!a). Interesting to now that Hugs seems to have somewhat laxer rules. Sep 2, 2017 at 20:06 • @Laikoni I'm looking at Haskell for months now and it doesn't cease to amaze me. I love the a=...:a trick. Thanks! Sep 3, 2017 at 6:32 • I don't know if you are aware of this but I think its worth mentioning. The reason Hugs is different here is that there is lower operator precedence for user defined operators than in the ghc. Sep 3, 2017 at 14:05 • @WheatWizard I wasn't aware. This makes perfect sense, given the error I got in GHC. Sep 3, 2017 at 17:32 Python 3, 87 85 bytes def f(s,l=65):c,*t=s;o=ord(c)-l;return o<0and'\n'+f(s)or' '*o+c+(t and f(t,o-~l)or'')  Try it online! J, 39 bytes (_65+3 u:[)}&(26{.'')/.~0+/\@,2>:/\3&u:  Try it online! Charcoal, 15 bytes Ｆθ«Ｊ⌕αι⁺ⅉ‹⌕αιⅈι  Try it online! Link is to verbose version of code. Explanation:  θ Input string Ｆ « Loop over characters α α Uppercase letters predefined variable ι ι Current character ⌕ ⌕ Find index ⅈ Current X co-ordinate ‹ Compare ⅉ Current Y co-ordinate ⁺ Sum Ｊ Jump to aboslute position ι Print current character  APL (Dyalog Classic), 20 bytes ⊃¨⎕a∘.∩⍨⊢⊂⍨1,2≥/⎕a⍳⊢  Try it online! K (ngn/k), 29 28 bytes {{x@x?c$65+!26}'(&~>':x)_x}


Try it online!

{ } function with argument x

>':x for each char, is it greater than the previous char?

~ negate

& where (at which indices) do we have true

( )_x cut x at those indices, return a list of strings

{ }' for each of those strings

c\$65+!26


the English alphabet

x? find the index of the first occurrence of each letter in x, use 0N (a special "null" value) if not found

x@ index x with that; indexing with 0N returns " ", so we get a length-26 string in which the letters from x are at their alphabetical positions and everything else is spaces

R, 129 117 bytes

function(s){z={}
y=diff(x<-utf8ToInt(s)-64)
z[diffinv(y+26*(y<0))+x[1]]=LETTERS[x]
z[is.na(z)]=" "
write(z,1,26,,"")}


Try it online!

Explanation (ungolfed):

function(s){
z <- c()                  # initialize an empty vector
x <- utf8ToInt(s)-64      # map to char code, map to range 1:26
y <- diff(x)              # successive differences of x
idx <- cumsum(c(          # indices into z: cumulative sum of:
x[1],                  # first element of x
ifelse(y<=0,y+26,y)))  # vectorized if: maps non-positive values to themselves + 26, positives to themselves
z[idx] <- LETTERS[x]      # put letters at indices
z[is.na(z)] <- " "        # replace NA with space
write(z,"",26,,"")        # write z as a matrix to STDOUT ("") with 26 columns and empty separator.