# Output the missing integer

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

• @riker That seems to be about finding a number missing in a sequence. This seems to be about finding a digit missing from a set. Commented Mar 27, 2017 at 19:48
• @Riker I wouldn't think it's a duplicate, given that the linked challenge has a strictly incrementing sequence (of potentially multi-digit numbers), whereas here it's in arbitrary order. Commented Mar 27, 2017 at 19:50
• Hi Josh! Since no one else has mentioned it so far, I'll direct you to the Sandbox where you can post future challenge ideas and get meaningful feedback before posting to main. That would have helped iron out any details (like STDIN/STDOUT) and resolved the duplicate dilemma before you received downvotes here. Commented Mar 27, 2017 at 20:23
• It's such a shame that 9-x%9 works for any digit except 0. Maybe someone more clever than me will find a way to make it work. Commented Mar 28, 2017 at 0:45
• Several answers take an integer as function input. Is that allowed? Commented Mar 28, 2017 at 0:52

# Python 2, 18 16 bytes

+beauty thanks to @Sarge Borsch

99066**2.strip


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

• 32043 can be changed to a more beautiful number. 99066 is center-symmetric (doesn't change if rotated 180 degrees around center) or maybe 97779 (palindrome, two distinct digits) Commented Mar 28, 2017 at 10:12
• If the OP allows to print the number twice, 764**4 can save two bytes. Commented Mar 30, 2017 at 9:16
• @Titus 764**4 is missing 5,8 and 9
– Rod
Commented Mar 30, 2017 at 11:15
• Typo ... I meant 763**4 = 338920744561 Commented Mar 30, 2017 at 12:00

# Python, 22 bytes

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


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

• Can you explain why this works? Commented Mar 28, 2017 at 20:07
• @KarlKastor, modulo 15 in base 16 works analogically to modulo 9 in base 10. Modulo base-1 is constant when taking sum of digits, just because 10 ≡ 1 (mod base-1). Sum of all possible digits is constant, so the missing digit is the difference of this constant and the input number (modulo base-1).
– mik
Commented Mar 29, 2017 at 13:43

# APL (Dyalog), 4 bytes

### Derived function

⎕D∘~


⎕DDigits

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

~ except [the argument]

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### Function train

⎕D~⊢


⎕DDigits

~ except

⊢ the right argument

Try it online!

### Explicit program

⎕D~⍞


⎕DDigits

~ except

⍞ character input

Try it online!

# 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! # Brain-Flak, 4838 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. • How does this work? Commented Mar 27, 2017 at 19:51 • @ГригорийПерельман Explanation added! Commented Mar 27, 2017 at 19:58 • If you move the negative(sum(input())) to the end, you can abuse the stack-height nilad to push 525 easier. (([][][]()()()){()()({}[()])}{}[{{}}]) should save you 10 bytes Commented Mar 27, 2017 at 20:04 • 30 bytes by subtracting 477 instead – Jo King Commented Apr 6, 2018 at 15:41 # Jelly, 3 bytes ØDḟ  Simply filters (ḟ) the input string from “0123456789” (ØD). Try it online! • I like the way that all the golfing languages (and even some non-golfing languages) are using the same algorithm, but Jelly manages to have the shortest names for the builtins it uses, and the least boilerplate for reversing the arguments to ḟ. – user62131 Commented Mar 27, 2017 at 20:26 • @ais523 APL is letter by letter the same (except that it would be an APL snippet rather than a function/program): Ø=⎕, D=D, ḟ=~, as in ⎕D~'867953120'. – Adám Commented Mar 27, 2017 at 21:20 • Me, while scrolling through the answers: "I'm predicting 3 characters in Jelly." Bingo. :^D Commented Mar 28, 2017 at 0:09 # 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"]  # 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> • s=>eval([1,...s].join^) saves a byte. – Neil Commented Mar 28, 2017 at 9:37 • @Neil ... and it's much more interesting too Commented Mar 28, 2017 at 10:05 • I feel this tip came suspiciously short after my answer :D anyway nice golf +1. Commented Mar 28, 2017 at 10:13 • @Christoph Well it sounded as if you wanted to spread the word... – Neil Commented Mar 28, 2017 at 10:49 • @Neil absolutly :) Nice to see that it helped! Commented Mar 28, 2017 at 10:51 # 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  • The first line of the second answer can be just .. – Neil Commented Mar 27, 2017 at 20:37 • You can save a few bytes by using deduplication instead of replacement: Try it online Commented Mar 27, 2017 at 20:39 • Oh well, and here was I just getting the second answer down to the old byte count of the first answer... ^ 5 ^. $*9¶ . $*_ +_¶_ ¶ _ – Neil Commented Mar 27, 2017 at 20:41 • @Neil I got it to one less than the original. Commented Mar 27, 2017 at 20:51 • (Huh, that looks suspiciously similar to my answer, the only difference is that you switched from _ to 1 to save a byte.) – Neil Commented Mar 27, 2017 at 20:56 # 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. # 05AB1E, 6 bytes 45¹SO-  Try it online! 45 # Push 45 ¹ # push input S # Split O # Sum - # Subtract (45 - sum)  # 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. ## JavaScript (ES6), 312928 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.

• s=>[...s].map(c=>r-=c,r=45)|r ;-) Commented Mar 27, 2017 at 21:01
• You are too much in love with reduce. +1 anyway Commented Mar 27, 2017 at 21:06
• @Arnauld I don't see that working when s[0]!='0', but there's already an answer that uses eval.
– Neil
Commented Mar 28, 2017 at 8:58
• Could you do s=>(15-0x${s}%15)%15? Commented Mar 28, 2017 at 14:40 • @Arnauld Bah, and I'd already done that for the Batch port too... – Neil Commented Mar 28, 2017 at 14:55 # Octave, 22 bytes @(x)setdiff('0':'9',x)  Try it online! # PHP, 27 <?=trim(32043**2,$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.

• Nice I have not think about trim. Alternatives values 32043,32286,33144,35172,35337,35757,35853,37176,37905,38772,39147,39336,40545,42744,43902,44016,45567,45624,46587,48852,49314,49353,50706,53976,54918,55446,55524,55581,55626,56532,57321,58413,58455,58554,59403,60984,61575,61866,62679,62961,63051,63129,65634,65637,66105,66276,67677,68763,68781,69513,71433,72621,75759,76047,76182,77346,78072,78453,80361,80445,81222,81945,83919,84648,85353,85743,85803,86073,87639,88623,89079,89145,89355,89523,90144,90153,90198,91248,91605,92214,94695,95154,96702,97779,98055,98802,99066 Commented Mar 28, 2017 at 10:43

## Bash + coreutils, 19 bytes

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

sum -s|dc -e524?--P


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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.

# 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


# 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


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• If you moved the space before the $1 it would be more obvious... – Neil Commented Mar 28, 2017 at 9:34 • @Neil, yep, that's a nice idea ! Thx ! Commented Mar 28, 2017 at 9:50 • You could use seq instead of echo: seq 0 9|tr -d \\n$1 Commented Mar 28, 2017 at 15:25

# 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:

• The number 2^29 has all the digits but 4, so 33 bytes: =REGEXEXTRACT(4&2^29,"[^"&A4&"]") Commented Mar 31, 2017 at 0:28
• @SteveKass Nice. =REGEXEXTRACT(0&49^9,"[^"&A1&"]") is also a valid solution, given similar logic. Updated answer. Commented Mar 31, 2017 at 22:25

# Bash, 36 30 bytes

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


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Posting to get golfing tips over this.

Thanks @DigitalTrauma, big fan ;)

fold -1 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.

• Without changing the approach, a couple of things to make this shorter: 1) have fold take input directly from STDIN instead of the herestring, and 2) have the (fold|seq) output pipe directly to sort, instead of the process substitution Commented Oct 17, 2022 at 17:22
• One of my first posts, I was an addict for some time... Commented Oct 19, 2022 at 14:36

# 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.

# CJam, 5 bytes

A,sq-


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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


# 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


# 05AB1E, 5 bytes

žhISK


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Explanation

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

• I like it because you can say it out loud. '"žhISK", he cried, as he waved his wand over the top-hat and a small white rabbit appeared in a puff of smoke'. Commented Mar 27, 2017 at 23:50
• SK can now be м to save a byte. Commented Dec 12, 2019 at 13:28

# Brachylog (2), 5 bytes

ẹ:Ị↔x


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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

• x implementation is old and pretty buggy, which is why you need ẹ. Commented Mar 28, 2017 at 6:23
• You can actually make an argument that something like ¬∋ℕ should work in only 3 characters – that's what I tried first – but there's multiple reasons why it doesn't, and I don't think there's any plausible way to change Brachylog so that it would.
– user62131
Commented Mar 28, 2017 at 15:15
• Having ¬∋ℕ work like that is not even possible in Prolog, unless specifically programming what you mean by not not in. ¬ in Brachylog is equivalent to \+ in Prolog, and its meaning is that of "not provable under the closed-world assumption", rather than "give me choice points for everything that does not verify this" (which is almost always an infinite number of things) Commented Mar 29, 2017 at 6:43
• The only way to do it in Prolog would be to "labelize" the ℕ in advance, but that would mean tinkering with Brachylog's evaluation order based on the content of the predicates. That's only one of the problems it has, though; there are a ton of others.
– user62131
Commented Mar 29, 2017 at 6:46

# Japt, 6 bytes

¬x n45


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### 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


# 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.

# 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.

# PHP, 37 bytes

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


# 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!