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.

Answer 1

Pyth - 7 bytes

+Qfql+Q

Try it online here.

Answer 2

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

Answer 3

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

Answer 4

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>

Answer 5

C, 67 65 61 bytes

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

Wandbox

Answer 6

Lua 5.2, 32 Bytes

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

Where the variable a is the input string.

Answer 7

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)

Answer 8

Python 2, 54 48 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

Answer 9

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 10

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 11

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 12

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.

Answer 13

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 14

Haskell, 61 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 15

Ruby, 62 58 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.

Answer 16

Perl 6,  46  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 17

05AB1E, 11 bytes

[¹¾JDg¾Q#¼\

Pretty straightforward bruteforce:

            Implicit i = 0
[           while true
 ¹¾J        Concatenate input and i -> str
    Dg¾Q#   Break if length(str) == i
         ¼\ Else, i += 1

Try it online!

Answer 18

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.

Answer 19

PHP, 42 bytes

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

Run with -r. Test at OnlinePHPfunctions.

Answer 20

bash, 47 bytes

 for((n=-1;${#s} != $n;));{ s=$1$[++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 21

><> (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 22

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 23

C#, 77 bytes

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

Answer 24

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 25

R, 49 bytes

cat(a<-scan(,""),(t<-nchar(a))+nchar(t+1),sep='')

Pretty straightforward solution.

Answer 26

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 ]

Answer 27

Clojure, 72 bytes

(defn f([s](f s 1))([s n](if(=(count(str s n))n)(str s n)(f s(inc n)))))

Answer 28

Wolfram, 56

#<>ToString@Nest[l+IntegerLength@#&,l=StringLength@#,2]&

Given l = StringLength[x] it appends l + IntegerLength[l + IntegerLength[l]] to x.

Answer 29

Labyrinth, 48 45 41 bytes

)"   10/{:@!
.,;: _ { _ ;
   })"}) 10-9!@

Try it online!

Saved 4 bytes thanked to @Martin Ender

Answer 30

ForceLang, 83 bytes

set s io.readln()
label 1
set n 1+n
set t s+n
if t.len=n
 io.write t
 exit()
goto 1

Answer 31

awk, 37 bytes

{FS="";r=$0;$0=r NF;$0=r NF;$0=r NF}1

That will work in any awk that splits lines into characters when FS is NULL e.g. GNU awk and most others. WIth other awks you'd have to use the length() function instead of NF to calculate the number of characters in the record, e.g. awk '{r=$0;$0=r length;$0=r length;$0=r length}1' file.

Breakdown:

FS=""    # split the record at every character so NF is a count of characters in the record
r=$0     # save the original record
$0=r NF  # append the original length to the original record and cause awk to recalculate NF
$0=r NF  # append the new length to the original record and cause awk to recalculate NF again
$0=r NF  # append the final length to the original record for output
1        # a true condition invoking awks default action of printing the current record

Example:

$ awk '{FS="";r=$0;$0=r NF;$0=r NF;$0=r NF}1' file
aaa4
1
aaaaaaaa9
aaaaaaaaa11
a13
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa101
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa102

Answer 32

Jelly, 12 bytes

Many thanks to Dennis for his help in writing this answer. Golfing suggestions welcome. Try it online!

D;³L=
LÇ1#;@

Explanation

LÇ1#;@  Main link. Argument: s
          Runs the helper link until a suitable x is found.
L       Yields the length of s.
 Ç1#    Calls the helper link until one match is found.
    ;@  Appends the matching x to s.

D;³L=  Helper link. Argument: x
         Appends x to s and checks if its length is equal to x.
D      Convert x from integer to decimal.
 ;³    Append to our original string.
   L=  Check if the length of this new string is equal to x.

Answer 33

Perl, 31 bytes

Just going through all the numbers until one fits:

perl -lpe '1while++$n<length$_.$n;$_.=$n'

Answer 34

Ruby, 43 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 35

C, 75 bytes

Yes, there's another C answer but this (the same method as my Factor answer) is shorter:

i,z;f(char*s){i=strlen(s);z=1+log10(i);sprintf(realloc(s,i+z)+i,"%d",i+z);}

Ungolfed:

i, z;

g (char* s){
  i = strlen(s);
  z = 1 + log10(i);
  s = realloc(s, i + z);
  sprintf(s + i, "%d", i + z);
}

ceil(log10(n) is the number of digits in base 10 n, but 1+ is shorter than ceil().

Then, resize the string so we don't get a segfault, and format the number after the string part.

Answer 36

Under review

Powershell V2.0, 64 bytes

$s=$args[0]
$s+[string]($s.length+([string]($s.length)).length)

Example

PS C:\Users\***\Downloads\golf> .\str-length.ps1 'This is my strin3333333333333333333333333333333333333333333333333333
33333333333333333333333333333333333333333g'
This is my strin333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333g113

Answer 37

Scala, 75 bytes

def a(b:String,c:Int=0):String=if((b+c).length>c)a(b,c+1)else b+c

Answer 38

awk, 27 bytes

$0=$0length($0length($0NF))

Test it:

$ awk -F '' '$0=$0length($0length($0NF))' file
7aaaaaa8
8aaaaaaa9
9aaaaaaaa11
97aaaaa10aaaaaaaa10aaaaaaaa10aaaaaaaa10aaaaaaaa10aaaaaaaa10aaaaaaaa10aaaaaaaa10aaaaaaaa10aaaaaaaa99
98aaaaaa10aaaaaaaa10aaaaaaaa10aaaaaaaa10aaaaaaaa10aaaaaaaa10aaaaaaaa10aaaaaaaa10aaaaaaaa10aaaaaaaa101

Answer 39

C# / CS Script 99 87 79 bytes

Saved a few bytes switching to a lambda. And a few more moving the try/catch out of the for loop.

F=a=>{int i=0,l;try{for(;;)l=(a+(a+a.Length).Length)[++i];}catch{}return a+i;};

Try it here

Expanded:

F => a
{
    // 'i' is for traversing the string
    // 'l' is just a placeholder
    int i = 0, l;

    // wrap the loop in a try/catch
    try
    {
        // loop forever
        for (;;)
            // assign to 'l' the char at position ++i
            // throws an IndexOutOfRangeException when it falls off the end
            l = (a + (a + a.Length).Length)[++i];
    }
    // empty catch block
    catch {}

    // returns a string concat of the input a and index i
    return a + i;
}

Answer 40

Java 7, 67 bytes

String y(String y){int i=0;for(;(y+i).length()!=i;i++);return y+i;}

Try it online!