How much do I have to write?

Question

Writing out numbers is among the Hello worlds of programming, often the numbers 1-10.

I want to write out many numbers! Many, Many numbers. But how many numbers do I have to write?

Task

Given an integer input, give a number as output that would give me the number of digits that would be in a string containing all integer numbers in the range from 0 to the input, inclusive. The negation identifier ("-") counts as a single character.

Example I/Os

Input: 8
Written out: 0,1,2,3,4,5,6,7,8
Output: 9

Input: 101
written out: 0,1,2,3....,99,100,101
Output: 196

Input: 102
written out: 0,1,2,3....,100,101,102
output: 199

Input -10
Written out: 0,-1,-2,-3,-4,-5,-6,-7,-8,-9,-10
output: 22

This is a code-golf. The lowest number of bytes wins!

Answer 1

05AB1E, 3 bytes

Code:

ÝJg

Uses the CP-1252 encoding. Try it online!

Explanation:

Ý     # Range [0 .. input]
 J    # Join into one string
  g   # Get the length of the string

Answer 2

Python 2, 55 46 bytes

lambda n:len(`range(abs(n)+1)`)+2*~n+3*n*(n<0)

Try it online!

Getting better.

Answer 3

Röda, 23 bytes

f x{[#([seq(0,x)]&"")]}

Try it online!

Explained:

f x{[#([seq(0,x)]&"")]}
f x{                  } /* Defines function f with parameter x. */
        seq(0,x)        /* Creates a stream of numbers from 0 to x. */
       [        ]       /* Creates an array. */
                 &""    /* Joins with "". */
     #(             )   /* Calculates the length of the resulting string. */
    [                ]  /* Returns the value. */

Answer 4

Python 2, 41 bytes

f=lambda n:len(`n`)+(n and f(n+cmp(0,n)))

Try it online!

Answer 5

Bash + OS X (BSD) utilities, 24 22 bytes

Thanks to @seshoumara for saving 2 bytes.

seq 0 $1|fold -1|wc -l

Test runs on Mac OS X:

$ for n in 8 101 102 -10 -1 0 1; do printf %6d $n; ./digitcount $n; done
     8       9
   101     196
   102     199
   -10      22
    -1       3
     0       1
     1       2

---

Here's a GNU version:

Bash + coreutils, 40 38 bytes

Again, 2 bytes saved thanks to @seshoumara.

(seq $1 0;seq 0 $1)|uniq|fold -1|wc -l

Try it online!

Answer 6

Python 2, 83 ,78 64 bytes

shortest version:

lambda x:sum(map(len,map(str,(range(0,x+cmp(x,.5),cmp(x,.5))))))

this version saved 5 bytes, thanks to @numbermaniac :

x=input()
print len(''.join(map(str,(range(x+1)if x>0 else range(0,x-1,-1)))))

Try it online!

this one I came up with on my own after that (same amount of bytes):

x=input()
print sum(map(len,map(str,(range(x+1)if x>0 else range(0,x-1,-1)))))

Try it online!

Answer 7

Java 7, 74 bytes _{(recursive - including second default parameter)}

int c(int r,int n){r+=(n+"").length();return n>0?c(r,n-1):n<0?c(r,n+1):r;}

Explanation (1):

int c(int r, int n){     // Recursive method with two integer parameters and integer return-type
                         // Parameter `r` is the previous result of this recursive method (starting at 0)
  r += (n+"").length();  //  Append the result with the current number's width
  return n > 0 ?         //  If the input is positive
     c(r, n-1)           //   Continue recursive method with n-1
    : n < 0 ?            //  Else if the input is negative
     c(r, n+1)           //   Continue recursive method with n+1
    ?                    //  Else (input is zero)
     r;                  //   Return the result
}                        // End of method

---

Java 7, 81 79 bytes _{(loop - single parameter)}

If having a default second parameter as 0 for this recursive approach isn't allowed for some reason, a for-loop like this could be used instead:

int d(int n){String r="x";for(int i=n;i!=0;i+=n<0?1:-1)r+=i;return r.length();}

Explanation (2)

int d(int n){                 // Method with integer parameter and integer return-type
  String r = "x";             //  Initial String (with length 1 so we won't have to +1 in the end)
  for(int i=n; i != 0;        //  Loop as long as the current number isn't 0
      i += n < 0 ? 1 : 1)     //   After every iteration of the loop: go to next number
    r += i;                   //   Append String with current number
                              //  End of loop (implicit / single-line body)
  return r.length();          //  Return the length of the String
}                             // End of method

---

Test code:

Try it here.

class M{
  static int c(int r,int n){r+=(n+"").length();return n>0?c(r,n-1):n<0?c(r,n+1):r;}

  static int d(int n){String r="x";for(int i=n;i!=0;i+=n<0?1:-1)r+=i;return r.length();}

  public static void main(String[] a){
    System.out.println(c(0, 8) + "\t" + d(8));
    System.out.println(c(0, 101) + "\t" + d(101));
    System.out.println(c(0, 102) + "\t" + d(102));
    System.out.println(c(0, -10) + "\t" + d(-10));
  }
}

Output:

9   9
196 196
199 199
22  22

Answer 8

Ruby, 20 26 29 bytes

->x{[*x..-1,0,*1..x]*''=~/$/}

Try it online!

Answer 9

RProgN 2, 5 bytes

n0R.L

Explination

n0R   # A Stack of all numbers between 0 and the input converted to a number.
   .L # The length of the stringification of this.

Simple solution, works like a charm.

Try it online!

Answer 10

Mathematica, 48 47 46 bytes

-1 byte thanks to Martin Ender!

StringLength[""<>ToString/@Range[0,#,Sign@#]]&

Anonymous function, taking the number as an argument.

Shorter solution by Greg Martin, 39 bytes

1-#~Min~0+Tr@IntegerLength@Range@Abs@#&

Answer 11

PHP, 59 60 bytes

Outgolfed by Roberto06 - https://codegolf.stackexchange.com/a/112536/38505

Thanks to roberto06 for noticing the previous version didn't work for negative numbers.

Simply builds an array of the numbers, puts it to a string, then counts the digits (and minus sign)

<?=preg_match_all("/\-|\d/",implode(",",range(0,$argv[1])));

Run example: php -f 112504.php 8

• 8 test case
• 101 test case
• 102 test case
• -10 test case

Answer 12

Haskell, 39 38 bytes

f 0=1
f n=length$show=<<[0..n]++[n..0]

Try it online! Edit: saved 1 byte thanks to @xnor!

Explanation:

In Haskell for numbers a and b [a..b] is the range from a to b in 1-increments or 1-decrements, depending on whether b is larger a. So for a positive n the first list in [0..n]++[n..0] is [0,1,2,...,n] and the second one is empty. For negative n the second range yields [0,-1,-2,...,n] and the first one is empty. However if n=0 both ranges yield the list [0], so the concatenation [0,0] would lead to a false result of 2. That's why 0 is handled as a special case.

The =<<-operator on a list is the same as concatMap, so each number is converted into a string by show and all those strings are concatenated in one long string of which the length is finally returned.

---

Before xnor's tip I used [0,signum n..n] instead of [0..n]++[n..0]. signum n is -1 for negative numbers, 0 for zero and 1 for positive numbers and a range of the form [a,b..c] builds the list of numbers from a to c with increment b. Thereby [0,signum n..n] builds the range [0,1,2,...,n] for positive n and [0,-1,-2,...,n] for negative n. For n=0 it would build the infinite list [0,0,0,...] so we need to handle 0 as a special case, too.

Answer 13

Brachylog, 5 bytes

⟦ṡᵐcl

Try it online!

Builds the range [0,input], converts each number to string, concatenates into a single string and returns the length of the result

Answer 14

PHP, 41 35 bytes

Saved 6 bytes thanks to user59178

Since ʰᵈ's answer was wrong for a negative input, I took it upon myself to build a new solution :

<?=strlen(join(range(0,$argv[1])));

This function :

• Builds an array from 0 to $argv[1] (aka the input)
• Implodes it with an empty character (i.e. transforms it to a string)
• Echoes the length of the string

Try it here!

Answer 15

R, 26 20 bytes

sum(nchar(0:scan()))

Very basic approach:

• Make a vector 0:x

• Count the characters in each value (will be coerced to a string automatically)

• Sum

Not sure if there are any tricks to cut down the function definition? 6 bytes saved thanks to Giuseppe, by taking input from stdin instead.

Answer 16

Lua, 52 bytes

t=0;for i=0,io.read()do t=t+#tostring(i)end;print(t)

Iterates through a for loop from 0 - input, converts the integer i to a string and adds the length of the string to t before printing t

Answer 17

QBIC, 25 bytes

:[0,a,sgn(a)|A=A+!b$]?_lA

Explanation:

:[0,a     Read 'a' from the cmd line, start a FOR loop from 0 to 'a'
,sgn(a)|  with incrementer set to -1 for negative ranges and 1 for positive ones
A=A+!b$   Add a string cast of each iteration (var 'b') to A$
]         NEXT
?_lA      Print the length of A$

Answer 18

MATL, 11 bytes

0hSZ}&:VXzn

Try it online!

0h           % Implicitly input n. Append a 0: gives array [n 0]
  S          % Sort
   Z}        % Split array: pushes 0, n or n, 0 according to the previous sorting
     &:      % Binary range: from 0 to n or from n to 0
       V     % Convert to string. Inserts spaces between numbers
        Xz   % Remove spaces
          n  % Length of string. Implicit display

Answer 19

Batch, 110 bytes

@set/a"n=%1,t=n>>31,n*=t|1,t=1-t*n,i=0
@for /l %%i in (0,1,9)do @set/a"t+=(i-n)*(i-n>>31),i=i*10+9
@echo %t%

Computes sum(min(0,abs(n)+1-10^k),k=0..9)+(n<0?1-n:1). (I only have to go up to 9 because of the limitations of Batch's integer arithmetic.)

Answer 20

Python 2, 68 bytes

def f(i,j=1):
 if i==0:print j
 else:j+=len(`i`);f((i-1,i+1)[i<0],j)

Try it online!

Longer than but different from other Python solutions. Defines a recursive function called as e.g. f(10)

Answer 21

PowerShell, 23 bytes

-join(0.."$args")|% Le*

Try it online! (will barf on TIO for very large (absolute) inputs)

Uses the .. range operator to construct a range from 0 to the input $args (cast as a string to convert from the input array). That's -joined together into a string (e.g., 01234) and then the Length is taken thereof. That is left on the pipeline and output is implicit.

Answer 22

Perl 6, 18 bytes

{chars [~] 0...$_}

Try it

Expanded:

{  # bare block lambda with implicit parameter ｢$_｣

  chars        # how many characters (digits + '-')
    [~]        # reduce using string concatenation operator &infix:<~>
      0 ... $_ # deductive sequence from 0 to the input
}

Answer 23

JavaScript, 50 bytes

Collaborated with @ETHproductions

n=>{for(t="";n;n<0?n++:n--)t+=n;alert(++t.length)}

Answer 24

Bash + GNU coreutils, 28

eval printf %s {0..$1}|wc -c

Try it online.

Answer 25

Retina, 28 bytes

\d+
$*
1
$`1¶
1+
$.&
^-?
0
.

Try it online!

Explanation

\d+
$*

Convert the number to unary, keeping the sign untouched.

1
$`1¶

Each 1 is replaced by everything up to itself plus a newline. With this we get a range from 1 to n if n was positive, from -1 to n with an additional - at the start if it was negative. All numbers are in unary and separed by newlines.

1+
$.&

Convert each sequence of ones to the corresponding decimal number.

^-?
0

Put a 0 at the start, replacing the extra - if it's there.

.

Count the number of (non-newline) characters.

Answer 26

Emacs, 20 bytes

C-x ( C-x C-k TAB C-x ) M-{input} C-x e C-x h M-=

The command itself is 20 keystrokes, but I need clarification on how this should be counted as bytes. I reasoned that counting each keystroke as 1 byte would be the most fair. The command above is written conventionally for easier reading.

Explanation

C-x (

Begin defining a keyboard macro.

C-x C-k TAB

Create a new macro counter. Writes 0 to the buffer; the counter's value is now 1.

C-x )

End keyboard macro definition.

M-{input} C-x e

After hitting META, type in your input number. The C-x e then executes the macro that many times.

C-x h

Set mark to beginning of buffer (which selects all of the text thus generated).

M-=

Run character count on the selected region. The number of characters will be printed in the minibuffer.

Example

Apologies for the terrible highlighting color. This is an example of using this command with the input 100. The output is in the minibuffer at the bottom of the screen.

Answer 27

Perl 5, 32 bytes

31 bytes of code + -p flag.

$\+=y///c for$_>0?0..$_:$_..0}{

Try it online!

Answer 28

C#, 77 73 bytes

-4 bytes thanks to @Kevin Cruijssen

Lambda function:

(r)=>{var l="";for(int i=0,s=r<0?-1:1;i!=r+s;i+=s)l+=i;return l.Length;};

Ungolfed and with test cases:

class P
{
    delegate int numbers(int e);
    static void Main()
    {
        numbers n = (r) =>
        {
            var l = ""; 
            for (int i = 0, s = r < 0 ? -1 : 1; i != r + s; i += s)
                l += i; 
            return l.Length;
        };
        System.Console.WriteLine(n(8));
        System.Console.WriteLine(n(101));
        System.Console.WriteLine(n(102));
        System.Console.WriteLine(n(-10));
        System.Console.ReadKey();
    }
}

Answer 29

JavaScript, 44 bytes

f=(n,i=1)=>n<0?f(-n)-n:n<i?1:n+1-i+f(n,i*10)

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

Answer 30

REXX, 56 bytes

arg n
l=0
do i=0 to n by sign(n)
  l=l+length(i)
  end
say l

Answer 31

Jelly, 6 bytes

r0Ṿ€FL

Try it online!

Jelly's surprisingly bad at this, mostly because it doesn't have a dedicated builtin for stringifying integers; it does have a builtin for stringifying anything, but that means we can't have it autovectorize over lists.

Explanation

r0Ṿ€FL
r0      All numbers from {the input} to 0, inclusive
   €    For each of those numbers numbers:
  Ṿ       Find its string representation
    F   Concatenate the resulting strings
     L  Take the length of the resulting string

Answer 32

JavaScript (Node.js), 37 bytes

f=n=>n?(n+"").length+f(n>0?n-1:n+1):1

Port of my Python answer.

Try it online!

Answer 33

Japt, 4 bytes

ò ¬l

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Explanation:

ò ¬l
ò     // Creates a range from [0...Input]
  ¬   // Joins the array into a string
   l  // Returns the length

Answer 34

SQL (PostgreSQL), 76 bytes

SELECT sum(length(a::text))FROM generate_series(least(0,$1),greatest(0,$1))a

This is an SQL function.

Answer 35

Java, 211 bytes

interface n{static void main(String[]a){int n=Integer.parseInt(a[0]),k=1,i;a[0]="";if(n<0)k=-1;for(i=0;i!=n+k;i+=k){a[0]+=i;if(i!=n)a[0]+=',';}System.out.print(a[0]);System.exit(a[0].replace(",","").length());}}

Takes number as first command line argument. Returns output as an error code:

Input: -10

0,-1,-2,-3,-4,-5,-6,-7,-8,-9,-10
Process finished with exit code 22

Ungolfed version with comments:

interface n {
    static void main(String[]a) {
        int n = Integer.parseInt(a[0]),                 // Input number 
        k = 1,                                          // For loop increment
        i;                                              // For loop counter
        a[0] = "";                                      // Empty string
        if(n < 0) k = -1;                               // If input number is negative loop increment is negative
        for(i=0; i != n + k; i += k) {                  // For loop
            a[0] += i;                                  // Append number to string
            if (i != n)                                 // If not the last number
                a[0] += ',';                            // Append comma
        }
        System.out.print(a[0]);                         // Output string
        System.exit(a[0].replace(",","").length());     // Return length of string without commas as error code
    }
}

Answer 36

K, 29 Bytes

I have a feeling there is a better way...

f:{#,/$((+;-)x<0).(0;!1+abs x)}

(+;-)x<0     ---> Call this A. If x is less that 0, return - otherwise return +
(0;!1+abs x) ---> Call this B. A list; (0;the numbers from 1 to the absolute value of x)
A.B          ---> Apply A to B, e.g1. (-).(0;0 1 2) ==> 0 -1 -2 e.g2 (+).(0;0 1 2) ==> 0 1 2
,/$x         ---> $tring x (x is numbers asc or desc) and flatten the list ,/
#x           ---> Finally count the flattened list

Some tests...

  f 8
9
  f 10
12
  f 101
196
  f 102
199
  f -10
22
  f -100
293

Answer 37

Clojure, 48 bytes

(count(apply str(range(min 0 %)(max 1(inc %)))))

---

(defn how-much-to-write? [n]
  ; Count the characters in the resulting string
  (count
    ; Basically "joins" the list with ""
    (apply str
      ; Creates an ascending/descending range, depending on the value of n
      (range (min 0 n) (max 1 (inc n))))))

Answer 38

Stacked, 25 bytes

:sign\0\abs|>*''join size

Try it online!

Explanation

:sign\0\abs|>*''join size    stack: (n)
:                            stack: (n n)
 sign                        stack: (n sign[n])
     \                       stack: (sign[n] n)
      0                      stack: (sign[n] n 0)
       \                     stack: (sign[n] 0 n)
        abs                  stack: (sign[n] 0 abs[n])
           |>                stack: (sign[n] range[0, abs[n]])
             *               negates the range if negative
              ''join         said range as a single string
                     size    the desired length

Answer 39

Scala 75 Bytes

(x:Int)=>{
  val? =Math.abs _
  (0 to?(x))./:("")((a,c)=>a+","+c* ?(x)/x).size-1
}

I'm not sure how I can improve it any further. The Math.abs is really really annoying. I think there is probably a better way to go about doing the negative stuff.

Answer 40

Javascript, 65 bytes

T=n=>{m=n>0?1:-1,o="";for(i=0;i!=n+m;i=i+m){o+=i}return o.length}

Answer 41

Swift, 72 bytes

let f={i in(min(i,0)...max(0,i)).reduce(0){$0+"\($1)".characters.count}}

The "characters.count" really kills the score.

Answer 42

Retina, 42 41 bytes

-(.+)
$*-¶$1
.+$
$*
\d|$
$%` 
(1*) 
$.1
.

Try it online!

Explanation

-(.+)
$*-¶$1

If the number is a negative, let's say -n (i.e. there is a - in the input), then replace the input with n -s, and n on the next line. In the end we just want to count the characters, and for -n as input, there will be exactly n - signs in the range.

.+$
$*

Convert the number at the end of string into unary.

\d|$
$%` 

Replace any digit character or the end of the string with all the text before it on the same line, plus a space. For a sequence of 1s, this results in  1 11 111 ... up until the original sequence. This builds the range from 0 to n, in unary.

(1*) 
$.1

Conversion back to decimal. Any series of 1s followed by a space is replaced with the number of 1s.

.

Count the matches of . and output it. . matches any single character except linefeed.

Answer 43

Pip, 13 12 bytes

#J:^0\,a|a,1

Takes input from command-line argument. Verify all test cases at Try it online!

Explanation

Some weird tricks in this one:

0\,a is the inclusive range from 0 to a.

If a is negative, we want to take the inclusive range from a to 0 instead. A naive approach would be a ternary expression with a<0, but we can do better. The problem is that something like (0,-11) is a perfectly valid Range object and is truthy, even though the range it represents contains no numbers¹. We want to convert it to an empty list (falsey).

To achieve this, we apply unary ^ to split the Range into characters. This operator works itemwise, so for example giving it 0\,10 would result in the list [[0] [1] [2] [3] [4] [5] [6] [7] [8] [9] [1 0]]. Splitting into characters won't make a difference to the result, since we're going to join everything together anyway. But crucially, it forces the conversion of the Range into a List. So an input of -10 now results in the empty list [].

Now since [] is falsey, we can take the logical Or of the above with a,1 (i.e. range(a,1), not including the upper bound) and get the elements we want.

Next, we need to join them together. Unfortunately, Join has a higher precedence than Or, so we would need parentheses. But another approach is to modify Join with the compute-and-assign meta-operator : (which is like the = in += in C-like languages, but more flexible). This lowers the precedence of the operator such that parentheses aren't needed. Assigning to an rvalue is a warning, but not an error.

Finally, we take the length (#) of the resulting string, which is then autoprinted.

_{¹ Such "backwards" ranges are used in string slicing, where the negative number means "the index 11 characters from the end."}

Answer 44

C++ 102 Bytes

int f(int n){int o=0;std::string s;for(n=n<0?o=-n:n;n>=0;)s+=std::to_string(n--);return s.length()+o;}

Ungolfed

int f(int n)
{
    int o=0;
    std::string s;
    for(n=n<0?o=-n:n;n>=0;)
        s+=std::to_string(n--);
    return s.length()+o;
}

Will add explanations.

Answer 45

Pyth, 6 5 bytes

ljk}0

Uses ais523's Jelly algorithm, except that the range starts from 0.

Thanks to RK. for -1.

Answer 46

CJam, 15 bytes

ri_g\z),f*:`:+,

Try it online!

Explanation:

ri_g\z),f*:`:+, e# Accepts an integer from STDIN.
ri              e# Get integer from STDIN.
  _g\           e# Store the integer's sign behind it.
     z)         e# Take the absolute value and increment it, to make an
                e# implicit range.
       ,        e# Create range [0..N].
        f*      e# Multiply every integer in the range with the sign we
                e# stored earlier, so as to include the - signs.
          :`    e# Map repr.
            :+  e# Concatenate.
              , e# Length.

Answer 47

PHP, 38 35 bytes

<?=strlen(join(range(0,$argv[1]))); // 35 bytes
<?=strlen(join("",range(0,$argv[1]))); // 38 bytes

Useage: php file.php 8

The join is a little used alias of implode(), and php can easily join numbers together as a string. Titus commented that the glue isn't needed in join() if you want a empty string.

Answer 48

Java 8, 80 bytes

Golfed:

n->{String r="";for(int i=0;i<=(n<0?-n:n);++i)r+=i;return r.length()-(n<0?n:0);}

Ungolfed:

public class HowMuchDoIHaveToWrite {

  public static void main(String[] args) {
    for (final int[] test : new int[][] { { 8, 9 }, { 101, 196 }, { 102, 199 }, { -10, 22 } }) {
      final int input = test[0];
      final int expected = test[1];
      final int actual = q(n -> {
        String r = "";
        for (int i = 0; i <= (n < 0 ? -n : n); ++i) {
          r += i;
        }
        return r.length() - (n < 0 ? n : 0);
      } , input);
      System.out.println("Input:    " + input);
      System.out.println("Expected: " + expected);
      System.out.println("Actual:   " + actual);
      System.out.println();
    }
  }

  private static int q(java.util.function.IntFunction<Integer> f, final int input) {
    return f.apply(input);
  }
}

Answer 49

J, 20 bytes

(#":i.1+|p)-(p+|p)%2

Answer 50

C#, 59 76 bytes

n=>string.Concat(System.Linq.Enumerable.Range(n>0?0:n,(n<0?-n:n)+1)).Length;

Longer than I wanted it to be because of the minus and Range takes a start and a count not start and end.

Answer 51

GolfScript, 33 bytes

~.abs),\.{.abs/}0if:S;{S*`,}%{+}*

Try it online!

Explanation:

~.abs),\.{.abs/}0if:S;{S*`,}%{+}* # Evals STDIN.
~                                 # Eval.
 .abs),\                          # Store an absolute value inclusive range behind the original integer.
        .{.abs/}0if:S;            # Store the integer's sign in S.
                      {S*`,}%     # After applying the input's sign, get the length of the string representation of every integer in the range.
                             {+}* # Sum.

Answer 52

Python, 63 bytes

lambda n:sum(map(len,map(str,range(*[[0,n-1,-1],[n+1]][n>0]))))

Answer 53

Bean, 44 bytes

xxd-style hexdump:

00000000: 2653 4da0 6280 40a0 7852 a043 ccd0 80cc  &SM b.@ xR CÌÐ.Ì
00000010: a043 8b23 0020 8001 8b53 a062 4da0 4382   C.#. ...S bM C.
00000020: 53d0 80a0 1f20 8048 2043 253a            SÐ. . .H C%:

Equivalent JavaScript (adapted from @Dennis) (45 bytes):

(f=_=>A?(A+"").length+f(A-=Math.sign(A)):1)()

Explanation

Declares a recursive IIFE (immediately invoked function expression) f, using pre-initialized variable A with numeric input to count down and incrementally calculate the amount of characters in the range of numbers.

Try the demo here