Make a number palindrome

Question

Write a function that takes a number as an argument and makes it a palindrome by appending minimum number of digits. The number will be at max of 100 digits.

Sample Inputs
12
122
232
2323
1012121
Sample Outputs
121
1221
232
23232
101212101

Answer 1

J, ^{50, 32} 26 characters!

f=:{.@(,"1(-:|.)\.#|.@}:\)

eg

f '12'
121 
f '232'
232   
f '2323'
23232   
f '1012121'
101212101     

How it works (by example)

y =: '1012121'

[\.y   NB. Sub lists of y
1012121
012121 
12121  
2121   
121    
21     
1 

|.\. y  NB> Reverses of sub lists of y
1212101
121210 
12121  
1212   
121    
12     
1  

([\. y) -:"1 (|. \. y) NB. Which of them are equal? (those are palindromes)
                       NB. ( -:"1 ) checks equality item by item
0 0 1 0 1 0 1

(-:  |.)\. y NB. Shortcut of the above
0 0 1 0 1 0 1

(0 0 1 0 1 0 1) # }:\y NB. Choose (#) the palindrome prefixes (\)
10    
1012  
101212

y, |.'10'   NB. Reverse and append the first prefix.
101212101

Answer 2

Perl, 32 chars

s/((.)(?1)\2|.?)$/$&.reverse$`/e

Needs Perl 5.10 or later for regex features, but no special command-line switch.

Sample use:

$ perl -pe 's/((.)(?1)\2|.?)$/$&.reverse$`/e' << EOT
> 12
> 232
> 2323
> 1012121
> EOT
121
232
23232
101212101

Uses Perl 5.10's recursive regex extensions to match the longest trailing palindrome as such:

m/
    (      # paren 1 - a palindrome is either:
      (.)  # paren 2 - a character
      (?1) # a palindrome as defined in paren 1
      \2   # the same character as in paren 2
    |        # or:
      .?   # a 0- or 1-character string
    )
    $      # at end of string
/x

It then replaces it with itself ($&) and appends whatever the string started with ($`), reversed.

Answer 3

Brachylog 2, 8 bytes, language postdates challenge

ẹ;AcB↔Bc

Try it online! The question asks for a function, so I provided one; the TIO link takes an argument that runs a function like a full program.

Explanation

ẹ;AcB↔Bc
ẹ          Split {the input} into digits
 ;Ac       Append {the shortest possible} list
    B↔B    to produce a palindrome
       c   then concatenate the resulting list of digits back into a number

Answer 4

Python, 88 chars

def f(x):
 x,y=list(str(x)),[]
 while x!=x[::-1]:y+=x.pop(0)
 return''.join(y+x+y[::-1])

Answer 5

Python (101 96)

edit: Shortened based on @gnibbler's solution

def p(n):s=str(n);r=s[::-1];l=len(s);return[int(s+r[l-i:])for i in range(l)if s[i:]==r[:l-i]][0] 

Original:

def p(n):
 s=str(n);r=s[::-1];l=len(s)
 for i in range(l):
  if s[i:]==r[:l-i]:return int(s+r[l-i:])

Answer 6

Python - 98 chars

Based on Hoa's answer :)

def p(n):s=str(n);r=s[::-1];l=len(s);return next(int(s+r[l-i:])for i in range(l)if s[i:]==r[:l-i])

Answer 7

Golfscript - 32 chars

{`:s-1%:r,,{s<r+..-1%=*}%{}?~}:f

Answer 8

Haskell, 85

Using the same algorithm as most everyone else:

import List
r=reverse
f s=s++(r.snd.head.filter g.zip(tails s)$inits s)
g(s,_)=s==r s

Examples from problem description:

*Main> map (f.show) [12,232,2323,1012121]
["121","232","23232","101212101"]

Answer 9

Ruby 1.9, 72 chars

f=->x{x=x.to_s.split'';99.times{|i|x.insert~i,x[i]if x!=x.reverse};x*''}

Answer 10

JavaScript (ES6), 145 126 chars

 p=a=>{S=x=>x.split``.reverse();for(s=String(a),i=0;i<s.length;i++)if(x=s+S(s.substring(0,i)).join``,x==S(x).join``)return x}

Commented:

function palindrome(n){
  s = String(n);
  for(i=0;i<s.length;i++)
  {
    x=s+s.substring(0,i).split("").reverse().join("") //take first n characters, reverse and append to the end
    if(x==x.split("").reverse().join("")) //is the number a palindrome?
      return x;
  }
}

Answer 11

Ruby, 70 bytes

f=->x{x=x.to_s.chars;99.times{|i|x.insert~i,x[i]if x!=x.reverse};x*''}

Try it online!

Based on YOU'S answer, with chars instead of .split'' to gain 2 chars. And i'm sure there's way to squeeze just a bit more ><

Answer 12

Java, 174 bytes

x->{Function<String,String>r=t->new StringBuilder(t).reverse().toString();String y=r.apply(x),z=x;int m=x.length();while(!z.equals(r.apply(z)))z=x+y.substring(--m);return z;}

Ungolfed:

x -> {
        Function<String, String> r = t -> new StringBuilder(t).reverse().toString();
        String y = r.apply(x), z=x;
        int m = x.length();
        while (!z.equals(r.apply(z))) z = x+y.substring(--m);
        return z;
    }

I have a feeling it could be a lot tighter but it's not immediately obvious to me how. The Function eats up a lot of space but I needed it in two places.

This works for any string, not just numbers, and it can be any length.

Answer 13

PHP, 64 Bytes

for(;strrev($p=$argn.strrev(substr($argn,0,$i++)))!=$p;);echo$p;

Try it online!