Appending String Lengths

Question

Challenge:

Given a string s on the characters a-z, A-Z, 0-9, append the length of s to itself, counting the additional character(s) in the length as part of the total length of s.

Input:

Just a string of arbitrary length (can be empty).

Output:

The same string, but with its length appended to the end. The characters that represent the length should also be counted as part of the length. In cases where there are multiple valid lengths to append, choose the smallest one possible (see test cases for examples).

Test Cases:

INPUT -> OUTPUT // Comment
aaa -> aaa4
 -> 1 // Empty string
aaaaaaaa -> aaaaaaaa9 // aaaaaaaa10 would also normally be valid, but violates using the smallest number rule mentioned above
aaaaaaaaa -> aaaaaaaaa11
a1 -> a13 // Input can contain numbers at the end of the string, you do not have to handle the fact that it looks like 13 rather than 3.
Longer test case(s):
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa -> aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa101
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa -> aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa102

Rules:

This is code-golf, so shortest code in bytes wins. Standard loopholes are forbidden. Submissions may be an entire program or a function, and you may either print the result to stdout or return it as a variable from a function.

• Martin Ender: What characters can appear in the input?
• Yodle: @MartinEnder Alphanumerical characters only, 0-9 and A-Z/a-z. So yes, you can have strings with numbers at the end. I'll add a test case for one.

Answer 1

JavaScript (ES6), 32 bytes

f=(s,n=0)=>(s+n)[n]?f(s,n+1):s+n

How it works

f = (s, n = 0) => // given a string 's' and starting with n = 0:
 (s + n)[n] ? // if the Nth character of (s + n) exists:
 f(s, n + 1) // try again with n + 1
 : // else
 s + n // return s + n

Starting with N=0, we test the Nth character (0-based) of the string made of the concatenation of the original input string and the decimal representation of N. We increment N until this character doesn't exist anymore.

Example:

N = 0 : abcdefghi0
 ^
N = 1 : abcdefghi1
 ^
N = 2 : abcdefghi2
 ^
...
N = 8 : abcdefghi8
 ^
N = 9 : abcdefghi9
 ^
N = 10 : abcdefghi10
 ^
N = 11 : abcdefghi11 -> success
 ^

Test cases

f=(s,n=0)=>(s+n)[n]?f(s,n+1):s+n
console.log(f("aaa")); // -> aaa4
console.log(f("")); // -> 1
console.log(f("aaaaaaaa")); // -> aaaaaaaa9
console.log(f("aaaaaaaaa")); // -> aaaaaaaaa11
console.log(f("a1")); // -> a13

• mbomb007: Wow, JS is much terser than Python for this.
• Gowtham: @Arnauld I cannot get my head around this. Do you mind explaining how this code works?

Answer 2

LaTeX, 108/171

\newcounter{c}\def\s#1#2]{\stepcounter{c}\def\t{#2}\ifx\empty\t\arabic{c}\else#1\s#2]\fi}\def\q[#1]{\s#10]}

\q[] //1

• pajonk: Whoa, I don't think I've ever seen a latex answer on ppcg before.

Answer 3

JavaScript (ES6), 37 bytes

f=(s,t=s,u=s+t.length)=>t==u?t:f(s,u)

<input oninput=o.textContent=f(this.value)><pre id=o>

• Prasanna: When I clicked on Run Code Snippet I get to see an error message. I have no knowledge of Javascript - I was just trying
• Neil: @Prasanna Works for me in Firefox; which browser are you using?
• Ismael Miguel: @Prasanna Works on latest Google Chrome. Are you sure you aren't using IE11 or older, Opera or anything that doesn't support ES6?
• Prasanna: I'm using an old good chrome (Version 48.0.2564.97). I will try this with IE too. Can't update my chrome - office security issues

Answer 4

Lua 5.2, 32 Bytes

a=arg[1]print(a..#a+#(''..#a+1))

Where the variable a is the input string.

Answer 5

C, (削除) 67 (削除ここまで) (削除) 65 (削除ここまで) 61 bytes

x;f(*v){printf("%d",(int)log10(x=-~printf(v))*-~(x%10>8)+x);}

Wandbox

• cat: Ohh, yeah, I shoulda printf'd... Anyways, congrats on having the shorter C solution :D +1

Answer 6

Pyth - 7 bytes

+Qfql+Q

Try it online here.

Answer 7

Mathematica, 57 bytes

#<>ToString[(a=Length@#)+(i=IntegerLength)[a+i@a]~Max~1]&

Unnamed function taking an array of characters as input and returning a string. Uses the fact that if a is the length of the input, then the number to append to the input is a plus the number of digits in (a + the length of a), rather than just a plus the number of digits of a. Unfortunately it wouldn't give the right answer for the empty-string input without the ~Max~1 special case.

Answer 8

Perl 6, (削除) 46 (削除ここまで) 35 bytes

{$_~(.chars,*.chars+.chars...{$^a==$^b})[*-1]}

{$_~(.chars,*.chars+.chars...*)[2]}

Try it

Expanded:

{ # bare block lambda with implicit parameter 「$_」
 $_ # the input
 ~ # concatenated with
 ( # sequence generator
 .chars, # the number of chars in 「$_」 (seed the generator)
 *\ # Whatever lambda input (represents previous value)
 .chars # number of chars in that
 + # plus
 .chars # the number of chars in 「$_」
 ... # keep doing that until
 * # indefinitely
 )[2] # get the value at index 2 of the sequence
}

Answer 9

Pyke, 8 bytes (old version)

.f+liq)+

Explanation:

.f ) - first where (i++)
 + - input + i
 l - len(^)
 iq - ^ == i
 + - input + ^

Try it here! (New version, 9 bytes)

• Luis Mendo: It always confuses me how buried the actual output is among warnings or other messages :-)
• Blue: I should really get around to fixing the web bug in the copy link that automatically disables the warnings switch

Answer 10

Python 2, (削除) 54 (削除ここまで) (削除) 48 (削除ここまで) 46 bytes

Simple solution. Recursion ended up being shorter.

f=lambda s,n=0:f(s,n+1)if(s+`n`)[n:]else s+`n`

Try it online

• xnor: I think you can do (s+`n`)[n:] for n<len(s+`n`).

Answer 11

Haskell, 46 bytes

f s=[l|i<-[0..],l<-[s++show i],length l==i]!!0

Usage example: f "aaaaaaaa" -> "aaaaaaaa9".

Simply try all numbers starting with 0 and take the first that fits.

Answer 12

Brachylog, 13 bytes

l<L$@:?rc.lL,

Try it online!

Explanation

Basically a description of the problem. It will try every value of L bigger than the length of the input until it finds one for which, when concatenated to the input, is the length of that concatenation.

l<L length(Input) < L
 L$@ Convert L to a string
 :?rc. The Output is the concatenation of the Input with L as string
 .lL, The length of the Output is L itself

Answer 13

Brainfuck, 258 bytes

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

The input must be terminated by a linefeed (LF). Only works for inputs with a length lesser than 256 (including the LF).

Try it online!

Explanation

# read first char and add one to cell #1
# the cell after the input will contain the length
,>+<
# subtract 10 to check for LF
----------
# while the input is not 10 (LF)
[
# restore the input to its original value
++++++++++
# add one to the length
>+
# cut and paste the length to the next cell, then read the input
[>+<-],
# subtract 10 to check for LF
----------
]
# for input abc, the tape here would be: a b c *0* 4
# rewind to the beginning of the input
<[<]>
# print the input string
[.>]>
# convert the length to ascii chars and output them
>>++++++++++<<[->+>-[>+>>]>[+[-<+>]>+>>]<<<<<<]>>[-]>>>++++++++++<[->-[>+>>]>[+[-
<+>]>+>>]<<<<<]>[-]>>[>++++++[-<++++++++>]<.<<+>+>[-]]<[<[->-<]++++++[->++++++++
<]>.[-]]<<++++++[-<++++++++>]<.[-]<<[-<+>]

Note: I used code from this SO answer to convert the length to ascii output; I hope this is acceptable on PPCG. This is my first Codegolf submission, and my second BF program. Feedback is welcomed.

• cat: This isn't valid then, it must pass all test cases
• Forcent Vintier: So supporting a length up to 999 would be enough?
• cat: The spec says "arbitrary length" which means "as long as your language is capable of handling or without running out of memory"
• Dennis: The brainfuck interpreter you're using has 8-bit cells, so as long as your algorithm works for strings of arbitrary length, it should be fine if it fails for strings of length 256 or higher. The C and JavaScript submissions will also fail once the strings get too long.
• Forcent Vintier: Thank you Dennis, I will modify my submission accordingly

Answer 14

Python, 39 bytes

lambda a:eval('a+str(len('*3+'a))))))')

Longer form:

lambda a:a+str(len(a+str(len(a+str(len(a))))))

Iteratively in Python 2 (41 bytes):

x=a=input();exec"x=a+`len(x)`;"*3;print x

Starting with x as the input string a, applies the transformation x -> a + str(len(x)) three times. I'm still not clear why three applications are needed to always reach the fixed point.

• Tom Viner: Why 3 times? First to append the length of text, second to adjust the length to include the number, third in case that adjustment added an extra digit.

Answer 15

Retina, 22 bytes

\G`
.
x
+r`\d*$
$._
x

Try it online!

Ah well, if it wasn't for digits appearing in the input, this would be merely 11 bytes:

+r`\d*$
$._

Answer 16

Ruby, (削除) 62 (削除ここまで) (削除) 58 (削除ここまで) 56 bytes

s=gets.chomp;p s+"#{(s+"#{(s+"#{s.size}").size}").size}"

Tested in irb.

There's probably a better way to do this, but this was the first thing I came up with. Any help in golfing would be appreciated.

edit: I realized my use of parentheses was excessive.

• DepressedDaniel: You only use l in one place. If you inline that, you will save 3 bytes l=;. But your solution will still be longer than mine ;)

Answer 17

05AB1E, 11 bytes

[13⁄4JDg3⁄4Q#1⁄4\

Pretty straightforward bruteforce:

 Implicit i = 0
[ while true
 13⁄4J Concatenate input and i -> str
 Dg3⁄4Q# Break if length(str) == i
 1⁄4\ Else, i += 1

Try it online!

Answer 18

PHP, 42 bytes

while(++$n<strlen($a=$argv[1].$n));echo$a;

Run with -r. Test at OnlinePHPfunctions.

Answer 19

bash, 47 bytes

 for((n=-1;${#s} != $n;));{ s=1ドル$[++n];};echo $s

Save this as a script, and pass the input string as an argument.

It's a brute force implementation: try each number in turn until you find one which works.

Answer 20

><> (Fish) 35 bytes

i:1+?!v:o
ln;v9l< >
*9+>:&)?!^1l&a

Takes input onto the stack, checks the length against values 9,99,999... and if the length is larger than add 1 to the stack length.

Answer 21

Haskell, (削除) 61 (削除ここまで) 60 bytes

e=length
l%n|s<-show$l+1,n>e s=s|m<-n+1=(l+1)%m
c s=s++e s%2

Try it online!

Recursive solution. Usage:

Prelude> c "aaaaaaaaa"
"aaaaaaaaa11"

Answer 22

Ruby, (削除) 43 (削除ここまで) 36 bytes

Anonymous function, abusing the $. global.

->s{$.+=1while(t=s+"#$.").size>$.;t}

Call it like this:

f = ->s{$.+=1while(t=s+"#$.").size>$.;t}
f[""] # => "1"
f["a"] # => "a2"
f["aaaaaaa"] # => "aaaaaaa8"
f["aaaaaaaa"] # => "aaaaaaaa9"
f["aaaaaaaaa"] # => "aaaaaaaaa11"

Answer 23

Vyxal, 6 bytes

LJL+LJ

Try it Online!

Append the length of the string, take the length of that, then append the new length to the original. This does work with digit overflows (9, 99, etc...).

• pacman256: There's probably something I can do with iteratively adding and appending lengths until they are the same....
• naffetS: The code looks especially funny if you change the + to an J, which works fine: LJLJLJ
• naffetS: You actually don't need the Ṡ flag.
• lyxal: 3(LJ for 4 bytes
• pacman256: oh..... but mine is funnier :p

Answer 24

Pip, 8 bytes

L3Ya.#yy

Try It Online!

Explanation

Inspired by Pacmanboss256's Vyxal solution.

 ;; y is empty string; a is first command-line argument (implicit)
L3 ;; Loop 3 times:
 a. ;; Concatenate a with
 #y ;; Length of y
 Y ;; Assign the result to y
 y ;; After the loop, output y

Some worked examples:

Input a "" "aaa" "aaaaaaaaa"
Original y "" "" ""
After loop 1 "0" "aaa0" "aaaaaaaaa0"
After loop 2 "1" "aaa4" "aaaaaaaaa10"
After loop 3 "1" "aaa4" "aaaaaaaaa11"

• pacman256: +1, neat loop setup

Answer 25

K (ngn/k), 31 bytes

{$[9>t:#x;x,1ドル+t;x,$#x,(1ドル+t)]}

Try it online!

+22 bytes thanks to Dominic van Essen for pointing out a not-working test case (that should have worked)

• Dominic van Essen: I think this fails for "aaaaaaaaa": it outputs "aaaaaaaaa10" which is 11 characters long...
• oeuf: @DominicvanEssen ohhhhhhhh... now I understand why.
• mkst: fails for the longer test cases (i.e. you dont handle strings of length 99+)
• mkst: {x,$#x,$#x,$#x} for 15 bytes i think?
• mkst: {{y,$#x}/4#,x} for 14 bytes, maybe golfable further?

Answer 26

Vyxal, 26 bits^v2 , 3.25 bytes

3(LJ

Try it Online!

Bitstring:

01001000011111010110110111

Finally posting my golfing suggestion as a separate answer. 3 times, len top and append.

• noodle person: Nice, syntax is very similar to the Thunno 2 equivalent answer
• lyxal: @noodleperson which is also very similar to the 05AB1E equivalent answer :p

Answer 27

C#, 77 bytes

n=>{int a=n.Length;int c=(a+1).ToString().Length-1;return(n+(n.Length+1+c));}

• Laikoni: I don't now C#, but couldn't you use return(n+(a+1+c)) as a=n.Length ?
• Laikoni: And also drop the -1 from int c=(a+1).ToString().Length-1 and the +1 from the return?
• Laikoni: Wait, does this handle the larger test cases correctly? It looks like it returns aa...a100 instead of aa...a101 for the 99a test case.

Answer 28

MATL, 11 bytes

`G@Vhtn@>]&

Try it online! Or verify all test cases.

` % Do...while
 G % Push input
 @ % Push iteration index (1-based)
 V % Convert number to string
 h % Concatenate horizontally
 t % Duplicate
 n % Get length of concatenated string
 @ % Push iteration index
 > % True if length of concatenated string exceeds iteration index
] % End. Run next iteration if top of stack is true; else exit loop
& % Specifiy that next function (implicit display) takes only one input
 % Implicitly display top of the stack. This is the concatenated string
 % that had a length equal to the iteration index

Answer 29

Ruby, 51 bytes (program)

Ruby, 49 bytes (function)

Program (last newline is not necessary and thus unscored):

x=gets.strip
i=0
i+=1 until(y=x+i.to_s).size==i
p y

Function (last newline is scored):

def f x
i=0
i+=1 until(y=x+i.to_s).size==i
y
end

Answer 30

Factor, 55 bytes

It's a walk in the park! I came up with this in my head as soon as I read the question.

[ dup length dup log10 ⌈ + >integer 10 >base append ]