Output the missing integer

Question

You will be given a string. It will contain 9 unique integers from 0-9. You must return the missing integer. The string will look like this:

123456789
 > 0

134567890
 > 2

867953120
 > 4

Answer 1

Python 2, 18 16 bytes

^{+beauty thanks to @Sarge Borsch}

`99066**2`.strip

Try it online!

99066**2 is just a shorter way to generate a string that contains 0~9

Answer 2

Python, 22 bytes

lambda s:-int(s,16)%15

Try it online!

An arithmetic solution. Interprets the input string as hex, negates it, and takes the result modulo 15.

Answer 3

APL (Dyalog), 4 bytes

Derived function

⎕D∘~

⎕D Digits

∘ (ties a left argument to the following dyadic function to create a monadic function)

~ except [the argument]

Try it online!

---

Function train

⎕D~⊢

⎕D Digits

~ except

⊢ the right argument

Try it online!

---

Explicit program

⎕D~⍞

⎕D Digits

~ except

⍞ character input

Try it online!

Answer 4

Brain-Flak, 48 38 36 + 3 = 39 bytes

10 bytes saved thanks to DJMcMayhem!

((([]())[]{}){()()({}[()])}{}[{{}}])

Try it online!

Explanation

The sum of all the digits in Ascii is 525. This program sums up the input and subtracts it from 525 to get the missing digit.

((([]())[]{}){()()({}[()])}{}      )

Will push 525. This takes advantage of the fact that we know there will be 9 elements of input to begin with. This means that [] evaluates to 9 which allows us to get to large numbers like 525 quickly.

Next we have the bit:

                             [{{}}]

which will sum up the inputs and subtract it from the total.

Answer 5

Haskell, 24 23 bytes

(477-).sum.map fromEnum

Try it online! Usage: (477-).sum.map fromEnum $ "123456890". 477 is the sum of the character codes of the digits 1 to 9, excluding 0. This anonymous function computes 477 minus the sum of all digit character codes to find the missing one.

Turning the char digits to ints is one byte longer:

(45-).sum.map(read.pure)
foldl(\a b->a-read[b])45

Try it online!

Answer 6

Jelly, 3 bytes

ØDḟ

Simply filters (ḟ) the input string from “0123456789” (ØD).

Try it online!

Answer 7

JavaScript (ES6), 26

Edit 1 byte save thx @Neil, with a much more smarter trick

Xoring all the values from 1 to 9 gives 1. Xor 1 one more time and the result is 0. So, if any single value is missing, the result will be the missing value.

s=>eval([1,...s].join`^`)

Test

f=s=>eval([1,...s].join`^`)

function go() {

  var i=I.value;
  O.textContent = f(i)
}  

go()

<input oninput='go()' value='012987653' id=I>
<pre id=O></pre>

Answer 8

Ruby, 14

Sums the ascii codes and subtracts from 48*9+45

->s{477-s.sum}

Use like this

f=->s{477-s.sum}

puts f["123456789"]

Answer 9

Retina, 27 21 19 bytes

-6 Thanks to Basic Sunset
-2 Thanks to Martin Ender

.
$*_5$*
+`_1|1_

1

Try it online!

Replace every digit with that many _s and 5 1s:

.
$*_5$*

Remove all of the _s and a 1 for each:

+`_1|1_ 

Count the number of 1s left:

1

Answer 10

05AB1E, 6 bytes

45¹SO-

Try it online!

45     # Push 45
  ¹    # push input
   S   # Split
    O  # Sum
     - # Subtract (45 - sum)

Answer 11

JavaScript (ES6), 31 29 28 22 bytes

s=>(15-`0x${s}`%15)%15

Port of @xnor's Python answer, except that JavaScript only has a remainder operator rather than a modulo operator, so I can't do it in a single step. Edit: Saved 6 bytes thanks to @Arnauld.

Answer 12

Brainfuck, 17 15 bytes

-[-[->-<],]>++.

Try it out here. This solution works on standard Brainfuck (8-bit cells) only, as it relies on wrapping.

It's a rare day when Brainfuck can actually compete, but this challenge just happened to line up with the BF spec pretty well!

Instead of straight-up breaking down this answer, I'd like to step through the iterations I took, because I think it would be more understandable (and more interesting).
Note: this solution is inspired largely by Wheat Wizard's Brain-Flak answer.

Explanation

Step 1, 26 bytes

In his answer, Wheat Wizard pointed out that the sum of the ASCII values from 0-9 sum to 525. And since standard Brainfuck only has a notion of [0,255], this makes the value 525 % 256 = 13. That is to say, subtracting the ASCII values of the input from 13 nets you the missing digit.

The first version of this program was:
1. Put 13 in the first cell
2. Take inputs into the second cell
3. Subtract the second cell from the first cell
4. Jump to 2 if there are inputs remaining
5. Print the first cell

And here's the code for the simple solution:

+++++++++++++ #Set the first cell to 13  
>,            #Take inputs into the second cell  
[[<->-],]     #Subtract the second cell from the first cell and repeat until inputs are over  
<.            #Print the first cell  

Step 2, 19 bytes

As pointed out in his answer, since we know the input will be exactly length 9, we can use that value as a constant, and eliminate that long string of +'s right at the beginning.
It also doesn't matter at what point we add 13 (thanks, commutative property!), so we'll mix it in with the subtraction and printing steps.

,        #Take input to enter the loop
[[->-<], #Subtract the first cell from the second cell 
>+<]     #Add 1 for each input; totaling 9
>++++    #Add the missing 4 to make 13
.        #And print

This was my original answer to this problem, but we can do better.

Step 3, 17 bytes

Interestingly enough, the previous answer works even if we begin with a + instead of a ,

+[[->-<],>+<]>++++.

Brainfuck required something in a cell in order to begin a loop. We naively added that extra 4 in the end, when it could have gone in other places.

-[[->-<],>+<]>++.

With some totally intentional (read: trial and error) loop trickery, starting off the program with a - leads to two interesting results:

1. One gets added to the second cell (saving 1 byte at the end).
2. The loops runs one extra time, totaling 10 instead of 9 (saving another 1 byte).

1 + 10 + 2 = 13, and we end up with the original answer.

Looking back on it, this is probably an excessive write-up for such a simple Brainfuck program.

Step 4, 15 bytes

After thinking about this solution a bit more, I was able to cut off 2 bytes.

I wanted to clarify something about the previous step:
The minus to enter the loop effectively adds 1, but what it's actually doing is subtracting 255 from the second cell (resulting in 1).

It's obvious in retrospect, but subtracting 1 from the first cell is the same as adding 1 to the second cell (because everything in the first cell gets subtracted from the second cell.)

-[-[->-<],]>++.

I was able to remove the ">+<" by adding a "-" at the beginning of the first loop. It has to go there, and not where the ">+<" was, because the program will loop infinitely otherwise.

Answer 13

Octave, 22 bytes

@(x)setdiff('0':'9',x)

Try it online!

Answer 14

Mathematica, 25 bytes

477-Tr@ToCharacterCode@#&

Pure function taking a string as input and returning an integer. Mathematica has long command names and is reluctant to convert between strings and integers, which makes it particularly bad at this challenge. The best I could find was the algorithm from Level River St's Ruby answer, which does a computation based on the total of the ASCII codes of the input string; in Mathematica, this uses only one long command name.

Answer 15

PHP, 27

<?=trim(32043**2,$argv[1]);

uses the trick from Rod's answer to generate a string containing all digits then removes all digits except for the missing one.

---

PHP, 41

for($b=1;$i<9;$b^=$argv[1][$i++]);echo$b;

This one uses xor because I haven't seen it yet.

Answer 16

Bash + coreutils, 19 bytes

I found a shorter bash solution, that uses an interesting checksum approach:

sum -s|dc -e524?--P

Try it online!

Explanation:

The sum command prints a checksum and a block count. I don't know many details, but using the option -s (System V algorithm) will make the checksum equal to the ASCII sum of each input character code. As such, the checksum remains constant when the order of the same input characters changes.

Given 867953120 as test case (last example), here is how the script works:

• sum -s outputs 473 1. If no integer was missing, the checksum would have been 525.
• dc -e524? pushes 524 and then the pipe input. The stack is: 1 473 524. The idea is to subtract the checksum from 525, but since sum outputs 1 as well, I need to work with it.
• --P. After applying the two subtractions (524-(473-1)), the stack is: 52. With 'P' I print the character with that ASCII code: 4, the missing digit.

Answer 17

Fortran 95, 146 128 bytes

function m(s)
character(len=10)::s,t
t='0123456789'
do j=1,10
k=0
do i=1,9
if(s(i:i)==t(j:j))k=1
end do
if(k==0)m=j-1
end do
end

Not very short, I'm afraid.

Ungolfed:

integer function m(s)
    implicit none

    character(len=9)::s
    character(len=10)::t
    integer:: i, j, k

    t='0123456789'
    do j=1,10
        k=0
        do i=1,9
            if (s(i:i) == t(j:j)) k=1
        end do
        if (k==0) m=j-1
    end do

end function m

Answer 18

CJam, 5 bytes

A,sq-

Try it online!

A,     e# The range from 0 to 9: [0 1 2 3 4 5 6 7 8 9]
  s    e# Cast to a string: "0123456789"
   q   e# The input
    -  e# Remove all characters from the range that are in the input
       e# Implicit output

Answer 19

GNU sed, 36 bytes

Includes +1 for -r

s/$/0123456789/
:
s/(.)(.*)\1/\2/
t

Try it online!

s/$/0123456789/ # Append 0123456789
:               # Start loop
s/(.)(.*)\1/\2/ # remove a duplicate character
t               # loop if something changed

Answer 20

Brachylog (2), 5 bytes

ẹ:Ị↔x

Try it online!

Arguably should be shorter (I'm still confused as to why the ẹ is necessary), but this is the best I could do.

Explanation

ẹ:Ị↔x
ẹ      Split the input into a list of characters
 :Ị    Pair that list with the string "0123456789"
   ↔x  Remove all elements of the list from the string

Answer 21

Befunge 98, 14 12 bytes

I saved 1 byte by moving the program onto 1 line and 1 byte by doing some better math

~+;@.%a--7;#

Try it online!

Explanation

The sum of the ASCII values range from 477 to 468 depending on which number is missing. By subtracting this from 7, we get the range -470 to -461. By modding this number by 10, we get the range 0 - 9, which we can then print.

~+;       ;#    Sums the ASCII values of all characters to stdIn
~          #    The # doesn't skip over the ~ because it's on the end of a line
~               Once EOF is hit, the ~ reverses the IP's direction
          ;#    Jump the ; that was used before
       --7      Subtract the sum from 7 (really just 0 - (sum - 7))
     %a         Mod it by 10
   @.           Print and exit

The reason I use the ASCII values instead of taking integer input is because the & command in Try it Online halts on EOF (Even though it should reverse the IP). The ~ works correctly, though.

Old Program, 14 bytes

#v~+
@>'i5*--,

The sum of the ASCII values of all 10 digits is 525. By subtracting the sum of the given digits from 525, we get the ASCII value of the missing character.

#v~+         Sums the ASCII values of all characters on stdIn
             Moves to the next line when this is done
 >'i5*       Pushes 525 (105 * 5)
      --     Subtracts the sum from 525
@       ,    Prints and exits

Answer 22

Common Lisp, 47 bytes

(lambda(s)(- 45(reduce'+ s :key'digit-char-p)))

Ungolfed:

(lambda (s) (- 45 (reduce '+ s :key 'digit-char-p)))

Explaination:

(reduce '+ s :key 'digit-char-p)

This loops through the chars in s, converts them to digits, and adds them. Digit-char-p, conveniently, return the number of the char as its "true" value, so it can be used as a test or a conversion.

(- 45 ...)

Subtract from 45 gives back the digit that was missing from the input.

Answer 23

Cubix, 18 Bytes

5v&;52/ni?@.>!&oW+

Expanded

    5 v
    & ;
5 2 / n i ? @ .
> ! & o W + . .
    . .
    . .

Try it here

Uses the same sort of method as this brain-flak answer.

Create the value -525 on the stack by pushing 5, 2, concatenate, push 5, concatenate and negate.
Then repeatably get input and add until end of input is hit.
Remove the last input, negate(make positive) the last add result, output the character and halt.

The reason for working from -525 up is that the character output is hit for each input iteration. Since the value is negative, nothing is output until the loop is exited and the negative value is made positive.

Answer 24

PHP, 37 bytes

<?=45-array_sum(str_split($argv[1]));

Answer 25

Google Sheets, 39 33 bytes

Input is entered into cell A1.

Code:

=REGEXEXTRACT(4&2^29,"[^"&A1&"]")

Saved 6 bytes thanks to Steve Kass.

Previous Code:

=REGEXEXTRACT("0123456789","[^"&A1&"]")

Result:

Answer 26

Bash (+utilities), 22, 19 bytes

• Use seq instead of brace expansion, -3 bytes (Thx @Riley !)

seq 0 9|tr -d \\n$1 

Test

$seq 0 9|tr -d \\n123456789
0

Try It Online!

Answer 27

PowerShell, 30 bytes

param($n)0..9|?{$n-notmatch$_}

Try it online!

Takes input $n, constructs a range 0..9 (i.e., 0, 1, 2 ... 9), then uses a Where-Object clause (the |?{...}) to pull out the number that does regex -notmatch. That's left on the pipeline, output is implicit.

Answer 28

Pyth, 5 Bytes

-jkUT

try it!

explanation

-jkUT
    T   # 10
   U    # The unary range of ten: [0,1,..,9]
 jk     # join that on the empty string
-       # set minus

"-jUT" also kinda works but produces newlines for every int.

Answer 29

Röda, 28 bytes

{(_/"")|ord _|sum|chr 525-_}

Try it online!

Answer 30

05AB1E, 5 bytes

žhISK

Try it online!

Explanation

žh     # from the string "0123456789"
    K  # remove
  IS   # each digit of the input

Answer 31

Japt, 6 bytes

¬x n45

Try it online!

Explanation:

¬x n45
   n45     // 45-
¬          //    Split the input into an array "123" → ["1","2","3"]
 x         //    Return the sum of all the items ["1","2","3"] → 6
           // 45 - 6 = output

Answer 32

C (tcc), 36 31 bytes

f(long*s){s=9-(*s+s[1]%16)%15;}

Takes a string as input and returns an int. As written, this work only on little-endian architectures.

The lack of a return statement is undefined behavior, but this works with tcc and gcc.

Try it online!

Alternate version, 33 bytes, no UB

f(long*s){*s=57-(*s+s[1]%16)%15;}

Takes a string pointer as input and overwrites the string with the result (allowed by default).

While this is perfectly valid C, it will not work with compilers such as gcc, which store strings in read-only memory sections.

Try it online!

Answer 33

dc, 12 10 bytes

Cdi?B%-B%p

Try the dc version online!

This uses the fact that the sum of the digits from 0 to 9 is 45, which is 1 more than a multiple of 11.

The program works by viewing the input as a base 12 number, finding its remainder when divided by 11, and subtracting that from 12 (to find the missing digit). The only catch is that if 0 or 1 is the missing digit, this would give an answer of 11 or 12, respectively, so I mod out by 11 one additional time at the end to take care of those cases.

---

This yields a short bash solution also:

Bash + Unix utilities, 17 15 bytes

dc -eCdi?B%-B%p

Try the bash version online!

Answer 34

J, 9 bytes

Num_j_&-.

Try it online!

Answer 35

Perl 5, 21 bytes

20 bytes of code + -p flag.

s/./+$&/g;$_=45-eval

Try it online!
Note that the input needs to be supplied without final newline (with echo -n for instance).

---

Some other (longer) approaches (all with -p flag):

$\=45;$\-=$_ for/./g}{

$_=9876543210=~s/[$_]//gr

for$@(0..9){$\=$@if!/$@/}}{

Answer 36

Batch, 83 76 52 bytes

@set/pn=
@cmd/cset/a(641670-0x%n:~,4%-0x%n:~4%)%%15

Takes input on STDIN. Uses @xnor's hex modulo 15 trick, except that a) Batch only has 32-bit integers, so I have to split the string into two b) Batch only does remainder, not modulo, so I have to subtrat the values from a large multiple of 15 first.

Answer 37

PHP, 43 Bytes

for(;strpos(_.$argv[1],48+$i++););echo$i-1;

PHP, 50 Bytes

<?=join(preg_grep("#[{$argv[1]}]#",range(0,9),1));

Answer 38

QBIC, 25 bytes

;{~instr(A,!a$)|a=a+1\_Xa

Explanation

;       Get the input as A$
{       Start an infinite DO loop
~instr  Test if A$ has an occurrence of a% cast to string       
(A,!a$) a starts out as 0. !..$ casts to string.
|a=a+1  If we did find an instance, tets for the next a
\_Xa    Else, quit, printing our missing number.

Answer 39

C, 44 bytes

j;f(char*s){for(j=477;*s;)j-=*s++;return j;}

Try it online!

Answer 40

PHP, 53 bytes

There allready was a array_sum and regex solution, wanted to provide another:

print_r(array_diff(range(0,9),str_split($argvs[1])));

A few bytes more, but as bonus it will provide all missing numbers.

Answer 41

MATL, 6 bytes

4Y2jX-

Try it at MATL Online

Explanation

4Y2     % Pre-defined literal for '0123456789'
j       % Grab input as a string
X-      % Compute the set difference between the two, yields the characters in 
        % '0123456789' that are missing in the input
        % Implicitly display the result

Answer 42

JavaScript, 26 bytes

v=>45-eval([...v].join`+`)

I'm not a big fan of eval, but it does the job. The sum of all digits 0-9 is 45. 45 minus the sum of the passed-in digits is the value of the missing digit.

Test

f=v=>45-eval([...v].join`+`)


function test() {

  var i=I.value;
  O.textContent = f(i)
}  

test()

<input oninput='test()' value='012987654' id=I>
<pre id=O></pre>

Answer 43

Scala, 9 bytes

477-_.sum

To use it, assign it to a variable:

val f:(String=>Int)=477-_.sum

_ is syntactix sugar for the arguments of a function, so this expands to x => 477 - x.sum, which will subtract the sum of the ascii codes of the input from 477.

Answer 44

J, 14 bytes

((i.10)-."."0)

Try it online!

J is always surprisingly inadequate for golfing :(

(                )  NB. Monadic fork: (f g h) x = (f x) g (h x)
  (i.10)            NB. Array of integers from 0 to 9
            "."0    NB. Digits of string (". = evaluate, "0 = each atom)
         -.         NB. Except

Answer 45

TXR Lisp, 28 bytes:

This is 28 bytes. It reads a line of digits and yields a string of the missing ones as the result value;

(diff"0123456789"(get-line))

This uses the awk macro to do read every line of input and print the missing digits. It only adds one byte to the length:

(awk((mf(diff"0123456789"))))

At the system prompt:

$ txr -P '(diff"0123456789"(get-line))'
135249
0678

$ txr -e '(awk((mf(diff"0123456789"))))'
123456789
0
234567890
1
012357698
4
13579
02468
^D

Answer 46

PHP, 50 39 Bytes

function x($y){echo(45-array_sum(str_split($y)));}

Jörg Hülsermann's answer prompted me to try the CLI method, Thanks.

echo 45-array_sum(str_split($argv[1]));

Test it at the command line with:

php -r 'echo 45-array_sum(str_split($argv[1]))."\n";' /'12346789'

Test it (The old version) here if you'd like.

Wheat Wizard's answer put me in the right direction and I got help from This Answer

Answer 47

Keg, 14 8 bytes

?⅀ȍ$-

Try it online! -6 bytes thanks to @A̲̲

Answer History

14 bytes

`0123456789`᠀-

Try it online!

Answer 48

Wren, 31 bytes

Fn.new{|x|"0123456789".trim(x)}

Try it online!

Explanation

Fn.new{|x|                      // New anonymous function with parameter x
          "0123456789"          // Declare all the numbers
                      .trim(x)  // Trim out everything included in x
                              } // The remaining number is the result

Answer 49

W, 13 7 bytes

'0'9.St

Explanation

'0'9.   % Define the range of string 0 to 9
     S  % Swap so that instructions are in the right order
        % a[0], "0...9" -> "0...9", a[0]
      t % Trim out everything in 0...9 that appears in a[0]
        % The result is the remaining number

W, 8 bytes

CJ525S-C

Explanation

C         % Convert every character to its code point form
 J        % Sum the list
  525S-   % Minus 525 from the value
       C  % Convert to character form
          % Implicit output

Answer 50

TI BASIC, 44 bytes

Prompt Str1
For(A,0,9
If not(inString(Str1,sub("0123456789",A+1,1
Disp A
End

Iterates over a string of each digit, checking if each digit is in the input string. If a digit is not in the input string, that digit is displayed. The sub( command is 1-indexed, so we add 1 during that step which helps us also avoid an extra byte looping to 10. The byte count is affected by the 2 byte tokens inString(, sub(, and Str1, if you rely on the input sitting in Ans before execution that would cut 5 bytes.

Answer 51

Lambdabot Haskell, 14 bytes

(['0'..'9']\\)

After realizing I could just make a list with the String "0123456789" instead of the integers [0,1,2,3,4,5,6,7,8,9], the step for converting the string to integers could be skipped.

Type is now String -> String (takes a string and returns a string with the correct number). Call like:

(['0'..'9']\\)"149263708"         -- Outputs "5"

Previous:

([0..9]\\).map digitToInt

Uses set difference, unlike the other Haskell answer, type String -> Int

Call like

([0..9]\\).map digitToInt$"123467890"  -- Outputs 5

Answer 52

Ruby, 25 bytes

->x{'0123456789'.tr x,''}

Not terribly competitive, I'm afraid

Answer 53

Pip, 5 bytes

5 bytes seems to be the number for all golfing languages that aren't Jelly.

,tDCa

Takes input as a command-line argument. Try it online!

Explanation

,t     Range(10)
  DC   From each element, delete all characters...
    a  ...that are in 1st cmdline arg
       Concatenate and print (implicit)

The result is actually a list like ["";"";"";"";4;"";"";"";"";""] (which is what you get if you put RP at the front of the code). Lists by default are concatenated together before printing, so all you will see is 4.

Answer 54

C (clang), 31 bytes

f(x){return x?f(x/10)-x%10:45;}

Strictly speaking, f(x) returns the sum of the missing digits. This lets me knock off a digit, get the sum of the missing digits, then add it back on.

Try it online!

Answer 55

Octave, 20 19 14 bytes

@(a)477-a*~~a'

Try it online!

Developed independently but the same as some other answers.

a'          transpose of array `a`
~~a'        convert all elements of a' to 1
a*~~a'      matrix multiplication of `a` with a column vector of 1s
            that is equivalent to sum(a)
477-a*~~a'  subtract from [477=sum('0123456789')-48]

Answer 56

Java 7, 61 54 bytes

long d(int n){return(15-Long.valueOf(n+"",16)%15)%15;}

Based on @Neil's JavaScript (ES6) answer (just like JavaScript, Java uses remainder for negative-modulo. Original port is from @xnor's Python answer)

Old answer (61 bytes):

String c(int n){return"0123456789".replaceAll("["+n+"]","");}

Try it here.

Answer 57

REXX, 36 bytes

#=0123456789
arg n
say verify(#,n)-1

Answer 58

AHK, 39 bytes

i=[%1%]
Send % RegExReplace(99066**2,i)

Again, the inherent confusion between the input variable 1 and the actual number 1 causes confusion in AHK functions. At least it only cost 3 bytes this time.

99066**2 = 99066^2 = 9814072356 (a trick I copied from Rod's answer)

Answer 59

Bash, 36 bytes

sort <(fold -1<<<$a;seq 0 9)|uniq -u

Try it online!

Posting to get golfing tips over this.

fold -1<<$a writes one char of input per line

seq 0 9 writes 0..9 one per line after this.

Those lines are fed to sort and filtered by uniq -u displaying only not duplicated lines.

Answer 60

Pure bash, 28

a=1234567890
echo ${a//[$1]}

Try it online.

Answer 61

C#, 21 bytes

z=>45-z.Sum(v=>v-'0')

The expression v-'0' takes the char value such as '3' and subtracts the char value for '0', leaving the value 3, essentially converting the character 3 into the integer 3.

45 is the sum of all the numbers 0 through 9. 45 - the sum of the passed-in integers yields the missing integer.

Sample code

public static void Main()
{
    Func<string, int> X = z=>45-z.Sum(v=>v-'0');

    Console.WriteLine(X("123456789"));
    Console.WriteLine(X("134567890"));
    Console.WriteLine(X("867953120"));
}

Test Here: https://dotnetfiddle.net/yanR5S

Answer 62

Chip, 64 bytes

Azv~a
##'Bzv~b
`^v##'Czv~c
*f|`^-##'Dzv~d
Z~'~e `^-##'
ZZZZZZZ~S

Try it online!

Chip is a 2D language that operates on individual bits in a byte stream. Each byte of input is broken down into its component bits for computation, then stitched back together for output.

This solution uses a rather simple algorithm, that is only a minor twist on that used by many of the other solutions:

1. Start with a value of 0x2:

    0+1+2+3+4+5+6+7+8+9 =  45 (0x2D)
                    ~45 = -46 (0xD2)    *
             -46 & 0x0F =   2  (0x2)
   

   ^{* I only do half of two's complement negation; I skip the increment at the end for a reason. I'll get back to this.}

2. Add each code point in the string, masked with 0x0F, ignoring carries.

3. After the ninth addition, the current value will be equal to -1 minus the missing digit, masked with 0x0F:

   Let's say we were missing the digit '4':

                 -1 - 4 =  -5  (0xB)
                  ~(-5) =   4  (0x4)    *
   

   ^{* Again, we only do half of negation. This instance cancels out the offset introduced by the previous negation.}

4. Prefix the value with 0x3, giving the codepoint for the missing digit:

             0x30 | 0x4 =  52 (0x34)
   

What portions of the code do each bit? Well, they're a bit mixed together, but I'll try to cover the highlights:

• The # elements do the addition described in step 2.
• The z elements near the adders carry over the current value to the next cycle so that we can have a running sum.
• The Z elements keep track of which digit of the input we are on, handling the condition in step 3. They also provide the signal used to initialize the adders to 0x2, as described in step 1.
• The S element suppresses all output until the Z elements disable it on the final digit. (If you are looking at the TIO, try deleting this element to see all the intermediate values. Also, adding the flag -v will print the actual code points to stderr.)
• The letters A through D are reading the 4 low bits of the input, and a through d write those same bits of the output.
• The letters e and f provide the value 0x30 to the output as described in step 4.

Answer 63

D, 24 bytes

writeln(477-readln.sum);

Just add up all characters in the given string (not terminated by a newline). The sum is s = 48 * 9 + (0 + 1 + 2 + ... + 9 - x) = 477 - x, therefore, x = 477 - s.

The calls make use of optional parentheses (-4 bytes), and .sum invocation uses uniform function call syntax (-1 more byte).

Try it online!

Answer 64

Q, 24 bytes

{1#iasc"0123456789"in x}

Gets a boolean vector between the expectation and input, then sorts the indices of the expectation by that boolean vector in ascending order, first value always being the missing element since boolean for it is set to 0 and we know that value=index in this case.

Answer 65

Perl 5, 16 15 bytes

Execution:

perl -pe '$_=/0/?9-$_%9:0'

perl -pe '$_=-hex($_)%15'

15 14 bytes of code + 1 byte for -p switch.

Explanation:

If 0 exists in the input, return 9 - modulo 9 of the input, otherwise return 0.

Interprets input as hex string, negates it, and calculates modulo 15 (based on xnor's Python answer).

Answer 66

awk, 18 bytes

{$0=/0/?9-$0%9:0}1

Explanation: If 0 exists in the input, return 9 - modulo 9 of the input, otherwise return 0. Implicit print.

Answer 67

Noodel, 8 bytes

ɲdFḶṡạĖ⁻

Try it:)

---

How it works

ɲdFḶṡạĖ⁻
         # Input implicitly pushed onto the stack.
ɲd       # Pushes the string "0123456789" onto the stack.

  FḶṡạĖ⁻ # Loops nine times removing digits from the string "0123456789" producing the missing number.
  F      # Pushes the string "F" onto the stack.
   Ḷ     # Consumes the string "F" and converts to a base 98 number producing 9 then loops the following that many times.
    ṡ    # Swaps the string "0123456789" with the input on the stack.
     ạ   # Gets the ith element from the input and pushes it onto the top of the stack.
      Ė  # Grabs the string "0123456789" at the bottom of the stack and puts it at the top.
       ⁻ # Removes the element pulled out of the input from the string.
         # Implicit end of the loop.

         # Implicit push to the screen, outputting the missing digit.

Answer 68

Racket 106 bytes:

(filter-not(λ(x)(member x(string->list(number->string n))))
(for/list((i(range 48 58)))(integer->char i)))

Ungolfed:

(define (f n)
  (filter-not
   (λ(x)
     (member x
             (string->list
              (number->string n))))
   (for/list((i(range 48 58)))(integer->char i))))

Testing:

(f 867953120)

Output:

'(#\4)