Find the Serialized Integer

Question

Task

Write a program that will take (as input) a positive integer. It will then count up from 0, appending each integer to a String, only continuing if the length of the String is less than the value of the input.

A serialized integer is defined as the fully-formed integer with the maximum value belonging to the String. By "fully-formed", the integer should have no missing digits (which would occur if the length constraint of the String is met).

The output of the program should be the serialized integer for its respective, positive input.

Rules

It's code golf, so the shortest answer (in bytes) wins!
Input will always be positive.
The output must be an integer in base-10 (decimal).
The program must be 0-indexed.

Example Input | Output

   5 | 4   (0 1 2 3 4              - Length of 5)
  11 | 9   (0 1 2 3 4 5 6 7 8 9 1  - Length of 11)
  12 | 10  (0 1 2 3 4 5 6 7 8 9 10 - Length of 12)
1024 | 377 (0 1 2 3 4 5 6 7 8 ...  - Length of 1024)

Note(s)

If you have any questions (or would like me to provide more examples), please comment!
Inspiration: https://stackoverflow.com/questions/45034478/how-do-i-calculate-the-maximum-serialized-integers-in-1024-length-limit

Adding quote marks to the string in the examples might make it easier to understand that it's a string. — isaacg, Jul 11, 2017 at 20:18
So the first N-1 digits of the Champernowne constant, with a 0 prepended? — user45941, Nov 14, 2017 at 18:19

Neil · Accepted Answer · 2017-07-11 17:32:18Z

8

JavaScript (ES6), 40 37 bytes

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

<input type=number min=1 value=1 oninput=o.textContent=f(this.value)><pre id=o>0

Edit: Saved 3 bytes with some help from @Arnauld.

edited Jul 11, 2017 at 17:32

answered Jul 11, 2017 at 14:55

Neil

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Emigna · Accepted Answer · 2017-07-11 15:37:10Z

6

05AB1E, 10 7 bytes

Idea to use prefixes from Jonathan's Jelly answer

LηJ€g›O

Try it online!

Explanation

L         # range [1 ... input]
 η        # prefixes
  J       # join each to string
   €g     # get length of each string
     ›    # input is greater than string length
      O   # sum

edited Jul 11, 2017 at 15:37

answered Jul 11, 2017 at 15:10

Emigna

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Felipe Nardi Batista · Accepted Answer · 2017-07-11 14:32:58Z

5

Python 2, 55 bytes

Recursive lambda port from @officialaimm's answer.

f=lambda k,s='',i=1:k>len(s+`i`)and f(k,s+`i`,i+1)or~-i

Try it online!

answered Jul 11, 2017 at 14:32

Felipe Nardi Batista

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ETHproductions · Accepted Answer · 2017-07-12 21:44:30Z

5

Japt, 13 bytes

1n@P±X l >U}a

Test it online!

Explanation

1n@ P± X l >U}a
1nX{P+=X l >U}a
                   Implicit: U = input integer, P = empty string
  X{         }a    Return the first integer X in [0, 1, 2, ...] that returns a truthy value:
    P+=X             Append X to P.
         l >U        Return P.length > U.
                   This returns the first integer that can't fit into the U-char string.
1n                 Subtract 1 from the result.
                   Implicit: output result of last expression

edited Jul 12, 2017 at 21:44

answered Jul 11, 2017 at 13:53

ETHproductions

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Jonathan Allan · Accepted Answer · 2021-05-09 20:53:25Z

5

Jelly, 11 10 9 5 bytes

ẈÄ<⁸S

A monadic link taking a positive integer and returning a non-negative integer.

Try it online!

How?

ẈÄ<⁸S - Link: positive integer, n
Ẉ     - length of (implicit decimal digits) of each of (implicit [1..n])
 Ä    - cumulative sums
   ⁸  - n
  <   - (cumulative sum) less than (n)?
    S - sum

edited May 9, 2021 at 20:53

answered Jul 11, 2017 at 14:21

Jonathan Allan

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Jörg Hülsermann · Accepted Answer · 2017-07-11 13:57:27Z

4

PHP, 56 bytes

for(;$argn>$l=strlen($r);)$r.=+$i++;echo$i-1-($argn<$l);

Try it online!

answered Jul 11, 2017 at 13:57

Jörg Hülsermann

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Laikoni · Accepted Answer · 2017-07-11 14:33:36Z

4

Haskell, 55 53 50 bytes

(n#x)a|l<-a++show x=last$x-1:[n#l$x+1|length l<=n]

Try it online!

Usage is (1024#"") 0

edited Jul 11, 2017 at 14:33

Laikoni

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answered Jul 11, 2017 at 13:54

H.PWiz

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0xffcourse · Accepted Answer · 2017-07-11 14:42:36Z

4

Python 2, 60 59 58 bytes

Thanks @Felipe for 2 bytes

i,j,k='',1,input()
while len(i+`j`)<k:i+=`j`;j+=1
print~-j

Try it online!

edited Jul 11, 2017 at 14:42

answered Jul 11, 2017 at 13:50

0xffcourse

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Community · Accepted Answer · 2020-06-17 09:04:33Z

4

Pyth, 8 7 bytes

tf<Q=+d

Try it online. Test suite.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jul 11, 2017 at 15:03

PurkkaKoodari

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\$\begingroup\$ I love it! Using slicing as a comparison is brilliant. \$\endgroup\$
– isaacg
Jul 11, 2017 at 20:20
\$\begingroup\$ @isaacg It's one of the nice golf features of Pyth(on). I got the idea when I saw Neil's answer (indexing instead of slicing, but same idea). < num seq was also very helpful. \$\endgroup\$
– PurkkaKoodari
Jul 11, 2017 at 20:25

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Sean · Accepted Answer · 2017-07-11 17:37:23Z

Perl 6, 36 bytes

{(0...^{([~] 0..$^a).comb>$_})[*-1]}

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0 ...^ {...} is the sequence of numbers from zero until one less than the number for which the code block in braces returns true. (... without the caret would return the first number for which the block returned true.)
[~] 0 .. $^a is the concatenation of numbers from 0 up to the current number $^a (the parameter to the code block).
.comb is a list of all of the characters (digits) in the concatenated string. Interpreted as a number, it evaluates to the length of the string. .chars would be more natural to use here, since it evaluates directly to the length of the string, but the name is one character longer.
$_ is the argument to the top-level function.
[*-1] selects the last element of the generated list.

steenbergh · Accepted Answer · 2017-07-11 16:17:55Z

QBIC, 34 bytes

{A=!p$~_lB+A|>:|_xp-1|\B=B+A]p=p+1

Explanation

{           DO infinitely
A=!p$       Set A$ to p cast to num
            Note that p starts out as 0.
~      >:   IF the input number is exceeded by
 _l   |     the length of
   B+A      A$ and B$ combined
_xp-1|      THEN QUIT, printing the last int successfully added to B$
            The _X operator quits, (possibly) printing something if followed by a-zA-Z
            _x is slightly different, it prints the expression between the operator _x and |
\B=B+A      ELSE add A$ to B$
]           END IF
p=p+1       Raise p, and rerun

xnor · Accepted Answer · 2017-07-11 20:11:49Z

2

Python 2, 44 bytes

f=lambda k,i=0:-2*(k<0)or-~f(k-len(`i`),i+1)

Try it online!

answered Jul 11, 2017 at 20:11

xnor

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Eelvex · Accepted Answer · 2017-07-12 01:40:22Z

2

J, 26 bytes

(>i:1:)([:+/\[:>.10^.1+i.)

((>i:1:)([:+/\[:>.10^.1+i.))"0 ] 5 11 12 1024 2000 20000 100000 1000000
4 9 10 377 702 5276 22221 185184

edited Jul 12, 2017 at 1:40

answered Jul 12, 2017 at 0:50

Eelvex

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Leaky Nun · Accepted Answer · 2017-07-14 06:50:40Z

2

R, 43 bytes

n=scan();sum(cumsum(floor(log10(1:n))+1)<n)

Try it online!

answered Jul 14, 2017 at 6:50

Leaky Nun

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Razetime · Accepted Answer · 2021-05-09 16:29:36Z

2

Husk, 8 bytes

#ȯ<¹ṁLŀḣ

Try it online!

About as hard as I could condense it. Pyth is really hard to beat here.

answered May 9, 2021 at 16:29

Razetime

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\$\begingroup\$ Hard to beat, but at least possible to match... \$\endgroup\$
– Dominic van Essen
May 9, 2021 at 20:53

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Dominic van Essen · Accepted Answer · 2021-05-09 20:52:29Z

1

Husk, 7 bytes

#<¹∫mLḣ

Try it online!

How?

      ḣ    # range from 1..input;
    mL     # get the lengths of string representations of each element;
   ∫       # get the cumulative sums;
#          # how many of these
 <¹        # are less than the input?

answered May 9, 2021 at 20:52

Dominic van Essen

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emanresu A · Accepted Answer · 2021-08-27 11:04:10Z

1

Vyxal `s`, 5 bytes

ʁvL¦>

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ʁ     # 0...n 
 vL   # Get length of each 
   ¦  # Cumulative sums
    > # (input) is greater than each?
      # (s flag) sum ToS at end

answered Aug 27, 2021 at 11:04

emanresu A

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coltim · Accepted Answer · 2021-08-27 18:16:03Z

1

K (ngn/k), 13 bytes

{(+\#'$!x)'x}

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Takes input n as x.

(...)'x run a binary search (returning the index of the largest value equal to or smaller than x) on the list generated by...
(+\#'$!)
- $!x stringify each value from 0..x-1
- #' count the number of characters in each stringified number
- +\ take the cumulative sums

answered Aug 27, 2021 at 18:16

coltim

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Felix Guo · Accepted Answer · 2017-07-12 23:15:01Z

0

WendyScript, 42 bytes

<<f=>(x){<<n=""#i:0->x{n+=i?n.size>=x/>i}}

f(1024) // returns 377

Try it online!

Ungolfed:

let f => (x) {
  let n = ""
  for i : 0->x { 
    n+=i
    if n.size >= x 
    ret i
  }
  ret
}

answered Jul 12, 2017 at 23:15

Felix Guo

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Community · Accepted Answer · 2020-06-17 09:04:33Z

0

PHP, 41 bytes

while(strlen($s.=+$i++)<=$argn);echo$i-2;

Try it online.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jul 12, 2017 at 23:16

Titus

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Community · Accepted Answer · 2020-06-17 09:04:33Z

Java 8, 64 bytes

n->{int i=0;for(String t="0";;t+=++i)if(t.length()>n)return~-i;}

Or slight alternatives with the same byte-count:

n->{int i=0;for(String t="";;t+=i++)if(t.length()>n)return i-2;}
n->{int i=-1;for(String t="";;t+=++i)if(t.length()>n)return~-i;}

Explanation:

Try it here.

n->{                  // Method with integer as both parameter and return-type
  int i=0;            //  Integer `i`, starting at 0
  for(String t="0";   //  String, starting at "0"
      ;               //  Loop indefinitely
       t+=++i)        //    After every iteration: append the String with `i+1`
                      //    by first increasing `i` by 1 with `++i`
    if(t.length()>n)  //   If the length of the String is larger than the input:
      return~-i;      //    Return `i-1`
                      //  End of loop (implicit / single-line body)
}                     // End of method

G B · Accepted Answer · 2017-11-14 07:09:11Z

0

Ruby, 39 bytes

->n{~-(0..n).find{|x|0>n-="#{x}"=~/$/}}

Try it online!

answered Nov 14, 2017 at 7:09

G B

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displayname · Accepted Answer · 2017-11-14 14:58:24Z

0

Ruby, 44 bytes

Inspired by Kevin Cruijssen's JAVA answer. -4 bytes thanks to G B.

->n{i,t=0,'';t+="#{i+=1}"while t.size<n;i-1}

edited Nov 14, 2017 at 14:58

answered Nov 13, 2017 at 16:31

displayname

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\$\begingroup\$ (i+=1;t+=i.to_s) is the same as t+="#{i+=1}", only 4 bytes longer \$\endgroup\$
– G B
Nov 14, 2017 at 7:11
\$\begingroup\$ And if you do that, you don't need the variable t anymore, you can subtract the size from n and then compare with 0. \$\endgroup\$
– G B
Nov 14, 2017 at 7:14

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Xcali · Accepted Answer · 2017-11-14 16:40:28Z

0

Perl 5, 31 + 1 (`-p`) = 32 bytes

$".=++$\while$_>=length$"}{$\--

Try it online!

answered Nov 14, 2017 at 16:40

Xcali

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Stack Exchange Network

Find the Serialized Integer

Task

Rules

Example Input | Output

Note(s)

24 Answers 24

JavaScript (ES6), 40 37 bytes

05AB1E, 10 7 bytes

Python 2, 55 bytes

Japt, 13 bytes

Explanation

Jelly, 11 10 9 5 bytes

How?

PHP, 56 bytes

Haskell, 55 53 50 bytes

Python 2, 60 59 58 bytes

Pyth, 8 7 bytes

Perl 6, 36 bytes

QBIC, 34 bytes

Explanation

Python 2, 44 bytes

J, 26 bytes

R, 43 bytes

Husk, 8 bytes

Husk, 7 bytes

Vyxal `s`, 5 bytes

K (ngn/k), 13 bytes

WendyScript, 42 bytes

Ungolfed:

PHP, 41 bytes

Java 8, 64 bytes

Ruby, 39 bytes

Ruby, 44 bytes

Perl 5, 31 + 1 (`-p`) = 32 bytes

Your Answer

Not the answer you're looking for? Browse other questions tagged
code-golf
string
integer
or ask your own question.

Hot Network Questions

Find the Serialized Integer

Task

Rules

Example Input | Output

Note(s)

24 Answers 24

JavaScript (ES6), 40 37 bytes

05AB1E, 10 7 bytes

Python 2, 55 bytes

Japt, 13 bytes

Explanation

Jelly, 11 10 9 5 bytes

How?

PHP, 56 bytes

Haskell, 55 53 50 bytes

Python 2, 60 59 58 bytes

Pyth, 8 7 bytes

Perl 6, 36 bytes

QBIC, 34 bytes

Explanation

Python 2, 44 bytes

J, 26 bytes

R, 43 bytes

Husk, 8 bytes

Husk, 7 bytes

Vyxal s, 5 bytes

K (ngn/k), 13 bytes

WendyScript, 42 bytes

Ungolfed:

PHP, 41 bytes

Java 8, 64 bytes

Ruby, 39 bytes

Ruby, 44 bytes

Perl 5, 31 + 1 (-p) = 32 bytes

Your Answer

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Not the answer you're looking for? Browse other questions tagged code-golfstringinteger or ask your own question.

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Hot Network Questions

Vyxal `s`, 5 bytes

Perl 5, 31 + 1 (`-p`) = 32 bytes

Not the answer you're looking for? Browse other questions tagged
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or ask your own question.