No one, please!

Question

I absolutely hate the digit 1. So, I need your help to convert numbers to their "proper forms".

Numbers in proper form never have two 1s in a row. 101 is okay, but 110 is hideous.

To convert, simply skip all the improper numbers and count normally. For instance...

1 -> 1
2 -> 2
...
10 -> 10
11 -> 12
12 -> 13
...
108 -> 109
109 -> 120
110 -> 121
111 -> 122

and so on.

Your program should take an integer and output it in proper form. This is code-golf, so shortest code in bytes wins.

Answer 1

Bash + GNU utils, 36

seq $1$1|grep -v 11|sed -n "$1{p;q}"

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Answer 2

Burlesque, 10 bytes

{11~[n!}FO

Older versions:

ro{11~[n!}f[

ro{Sh"11"~=n!}f[

Answer 3

Jelly, 8 bytes

1w11¬$#Ṫ

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Almost entirely printable ASCII! Takes input from STDIN

How it works

1w11¬$#Ṫ - Main link. No arguments
     $   - Group the previous 2 links into a monad f(k):
  11     -   Yield 11
 w       -   Yield 1 if 11 is a sublist of k's digit else 0
    ¬    -   Map 1 to 0 and 0 to 1
1     #  - Count up k = 1, 2, 3, ... until n integers return true under f(k)
       Ṫ - Return the last k

---

Jelly, 8 bytes

²wÐḟ11ị@

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An alternate approach, still only 8 bytes though. Relies on the assumption that the output is always less than the square of the input

How it works

²wÐḟ11ị@ - Main link. Takes n on the left
²        - n²
  Ðḟ     - Filter-false the range [1, 2, ..., n²] on:
    11   -   11
 w       -   Is a sublist of the digits
      ị@ - Take the n'th value in this list

Answer 4

JavaScript, 53 bytes

n=>[...Array(n*2).keys()].filter(a=>!/11/.test(a))[n]

Alternate (using comprehensions, same length):

n=>[for(i of Array(n*2).keys())if(!/11/.test(i))i][n]

Answer 5

Perl 5, 34 Bytes

Looping a counter and changing the occasional double-one.

map{$i++;$i=~s/11/12/}1..pop;say$i

Test

$ perl -M5.012 -e 'map{$i++;$i=~s/11/12/}1..pop;say$i' 111
$ 122

Answer 6

Python 2, 50

lambda n:[i for i in range(n*2)if'11'not in`i`][n]

An anonymous function that lists the numbers not containing 11 in order, and takes the nth one. The off-by-one error of zero-indexing cancels with the inclusion of 0 in the list.

In theory, this will fail for sufficiently high numbers where f(n)>2*n, but this shouldn't happen until n is at least 10**50.

---

51 bytes:

n=input();i=0
while n:i+=1;n-='11'not in`i`
print i

Counts up numbers i until the quota of n numbers without 11 is met.

A function is the same length because of the off-by-one corrections needed.

f=lambda n,i=0:n+1and f(n-('11'not in`i`),i+1)or~-i

Answer 7

Pyth, 13 11 bytes

e.f!}`hT`ZQ

Saved 2 bytes thanks to @FryAmTheEggman.

Live demo and test cases.

13-byte version

e.f!}"11"+ZkQ

Answer 8

Python 3 74

Still need a bit of golfing.

n=int(input())
c=0
for x in ' '*n:
 c+=1
 while'11'in str(c):c+=1
print(c)

It's pretty brute force right now.

Answer 9

JavaScript (ES6) 43 bytes

As an anonymous function

n=>eval('for(i=0;/11/.test(i)||n--;i++);i')

Note: the very simplest way would be 44:

n=>{for(i=0;/11/.test(i)||n--;i++);return i}

Test running the snippet below.

f=n=>eval('for(i=0;/11/.test(i)||n--;i++);i')

var o='',v=0,r,c;
for(r=0;r<30;r++)
{
  o += '<tr>';
  for(c=0;c<10;c++)
  {
  
    o += '<td>' +v + '⇨' + f(v) + '</td>';
    v++;
  }
  o += '</tr>'

}
O.innerHTML=o

<table id=O style='font-size:80%'></table>

Answer 10

Haskell, 51 bytes

([x|x<-[0..],notElem('1','1')$zip=<<tail$show x]!!)

Usage example: ([x|x<-[0..],notElem('1','1')$zip=<<tail$show x]!!) 110 -> 121.

How it works:

[x|x<-[0..]                                   ]    -- take all x starting with 0
           ,                                       -- where
                   ('1','1')                       -- the pair of two chars '1'
            notElem                                -- is not part of
                             zip=<<tail            -- the list of pairs of neighbor elements of
                                        show x     -- the string representation of x
                                               !!  -- take nth element, where n is the parameter

Answer 11

GolfScript, 29 24 18 bytes

~0\{{).`11`?)}do}*

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This is my first code golf, so this is probably the least effecient way of doing things

-5 bytes by not being dumb and using the increment symbol as well as just keeping the input on the stack and decrementing instead of wasting space saving it as a variable and comparing it, probably more things I could save on

-6 bytes the loop is fixed size and I don't need the input after the loop so I can just use repeat

Answer 12

MUMPS, 37 bytes

t(i) f j=1:1 s:j'[11 k=k+1 q:k=i
    q j

Pretty straightforward. The only "interesting" thing here is the construct j'[11 - '[ is the "does not contain" operator, so that "abc"'["ab" is false and "abc"'["cd" is true. Despite both operands of j'[11 being numbers, MUMPS remains unperturbed. It will happily autocoerce both operands to strings and move on with its life. Hooray!

(Incidentally, if you're okay with the program never terminating, we can shorten this to 35 bytes: t2(i) f j=1:1 s:j'[11 k=k+1 w:k=i j)

Answer 13

Brachylog, 15 bytes

∧{ℕ₁≜¬s11&}ᶠ↖?t

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∧{ℕ₁≜¬s11&}ᶠ↖?t
∧{ …      }ᶠ↖?t get the N'th output from …
  ℕ₁≜             try a number, starting with 1
     ¬s11         doesn't contain 11
         &        output that number

Answer 14

Raku, 24 bytes

{grep(?1^/11/,^∞)[$_]}

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grep(..., ^∞) filters the numbers from zero to infinity according to the given test object, and [$_] indexes into that sequence with the function argument, returning the number at that index.

/11/ matches any number whose decimal representation contains two consecutive one digits. The sense of the match can be inverted by writing True ^ /11/, where ^ creates an exclusive-or junction matcher that succeeds only if exactly one of its operands succeeds. True will always succeed, so the overall junction can only succeed if the /11/ fails. Finally, True can be expressed more concisely as ?1, that is, 1 coerced to a Boolean.

none(/11/) also works as a matcher, but is longer.

Answer 15

Husk, 11 10 bytes

Edit: -1 byte thanks to Razetime

!fö¬€ḋ3QdN

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Finds index of input among natural numbers filtered to exclude those with digit-sublists that include [1,1].

Answer 16

AWK, 33 bytes

{for(;$1;)$1-=!index(++b,11)}$0=b

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It works by running a loop that increments a counter blindly each iteration, but only decrements the control variable if the blind counter does NOT include the string 11. Once the loop exits, the blind counter is printed as the answer.

 {                            }     - code block run for each input line
  for(;$1;)                         - loop until $1 is 0
           $1-=!index(   ,11)       - decrement $1 if "b" doesn't include "11"
                      ++b           - blindly increment a counter
                               $0=b - set $0 to blind count, prints automatically

Answer 17

Julia 1.0, 44 42 bytes

x->(r=1:9x)[.!occursin.("11",repr.(r))][x]

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Answer 18

05AB1E (legacy), 7 6 bytes

µN11å≠

-1 byte switching to the legacy version of 05AB1E, where the index is output implicitly after a while-loop µ. In the new 05AB1E version we have to push the N explicitly, and it'll output the top of the stack after the while-loop instead (so the N would be NN or ND).

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Explanation:

µ       # Loop until the counter_variable is equal to the (implicit) input-integer:
        # (the counter_variable is 0 by default)
 N11å   #  Check if the 0-based loop-index `N` contains 11 as substring
     ≠  #  Invert this boolean (1→0; 0→1)
        #  (implicitly pop the top of the stack, and increase the counter_variable by
        #  1 if it's truthy)
        # (after which the index `N` is output implicitly as result)

Answer 19

Perl 5, 47 bytes

@_[$_]=++$i!~/11/?$i:redo for 1..<>;print$_[-1]

Answer 20

PHP, 43 bytes

while(preg_match('/11/',$i)){$i++;}print$i;

Answer 21

Perl 5 -p, 25 bytes

$\++while$\=~/11/||$_--}{

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Answer 22

Java (JDK), 76 bytes

p->{int c=1,s=0;while((c+s+"").indexOf("11")>=0&&++s>0||c++<p);return p+s;};

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How it works: Lambda function obviously, but mostly it uses short-circuit logic in the while loop boolean to increment the appropriate counter keeping track of the cumulative shift for each input and returning the input plus that shift.

Answer 23

Python 3, 64 bytes

[print(e) for e in map(str,range(int(input())+1))if not'11'in e]

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Answer 24

Excel, 64 bytes

=LET(x,SEQUENCE(2^20),INDEX(FILTER(x,ISERROR(FIND("11",x))),I7))

Works up to 1,000,579 -> 1,048,576

Answer 25

Pip, 11 bytes

LaW11N Ui_i

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Explanation

             i is 0; a is first command-line argument (implicit)
La           Loop (a) times:
  W           Loop while
   11N        11 is a substring of
       Ui     i, incremented:
         _     No-op
          i  Once the loop completes, output i

Answer 26

Japt, 9 8 bytes

È¥srB}iU

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È¥srB}iU     :Implicit input of integer U
È            :Function taking an integer as argument
 ¥           :  Test equality with
  s          :    Convert to string
   r         :    Remove all occurrences of
    B        :      11
     }       :End function
      iU     :Get the Uth integer that returns true

Answer 27

Vyxal, 7 bytes

11c¬)ȯt

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      t # nth
     ȯ  # integer
    )   # which
   ¬    # does not
  c     # contain
11      # 11

Answer 28

JavaScript, 37 bytes

f=(n,i=0)=>n?f(n-!/11/.test(++i),i):i

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