Find the highest unique digit

Question

Surprisingly we haven't had a simple "find the highest digit" challenge yet, but I think that's a little too trivial.

Given input of a non-negative integer, return the highest unique (ie not repeated) digit found in the integer. If there are no unique digits, your program can do anything (undefined behaviour), other than the numbers that are non unique in the input of course.

The input can be taken as a single integer, a string, or a list of digits.

Test cases

12         -> 2
0          -> 0
485902     -> 9
495902     -> 5
999999     -> Anything
999099     -> 0
1948710498 -> 7

This is code-golf so fewest bytes in each language wins!

Answer 1

05AB1E, 4 3 bytes

Saved 1 byte thanks to Mr. Xcoder notifying that a digit list is valid input.

¢ÏM

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Explanation

¢     # count occurrences of each digit in input
 Ï    # keep only the digits whose occurrences are true (1)
  M   # push the highest

Answer 2

Python 3, 40 bytes

Saved 2 bytes thanks to movatica.

lambda i:max(x*(i.count(x)<2)for x in i)

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42 bytes

Works for both String and list of digits parameter types. Throws an error for no unique digits, kind of abuses of that spec:

lambda i:max(x for x in i if i.count(x)<2)

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---

Explanation

• lambda i: - Declares a lambda function with a string or list of digits parameter i.
• max(...) - Finds the maximum value of the generator.
• x for x in i - Iterates through the characters / digits of i.
• if i.count(x)<2 - Checks if the digit is unique.

Answer 3

Alice, 15 bytes

/&.sDo
\i-.tN@/

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Explanation

/...
\.../

This is a simple framework for linear code that operates entirely in Ordinal mode (meaning this program works completely through string processing). The unfolded linear code is then just:

i..DN&-sto@

What it does:

i    Read all input as a string.
..   Make two copies.
D    Deduplicate the characters in the top copy.
N    Get the multiset complement of this deduplicated string in the input.
     This gives us a string that only contains repeated digits (with one
     copy less than the original, but the number of them doesn't matter).
&-   Fold string subtraction over this string, which means that each of
     the repeated digits is removed from the input.
s    Sort the remaining digits.
t    Split off the last digit.
o    Print it.
@    Terminate the program.

Answer 4

Retina, 16 bytes

O`.
(.)\1+

!`.$

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Explanation

O`.

Sort the digits.

(.)\1+

Remove repeated digits.

!`.$

Fetch the last (maximal) digit.

Answer 5

Charcoal, 18 12 bytes

Ｆχ¿⁼№θＩι¹ＰＩι

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Prints nothing if no solution is found. The trick is that the for loop prints every unique number in the input string, but without moving the cursor, thus the value keeps reprinting itself until the final solution is found.

The previous version printed the characters A to Z when no solution was found, hence the comments:

ＡααＦχＡ⎇⁼№θＩι¹Ｉιααα

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Answer 6

Husk, 7 bytes

→fo¬hgO

Try it online! (Test suite, crashes on the last test case since it has no unique digits)

This is a composition of functions in point-free style (the arguments are not mentioned explicitely anywhere). Takes input and returns output as a string, which in Husk is equivalent to a list of characters.

Explanation

Test case: "1948710498"

      O    Sort:                             "0114478899"
     g     Group consecutive equal elements: ["0","11","44","7","88","99"]
 fo¬h      Keep only those with length 1*:   ["0","7"]
→          Take the last element:            "7"

*The check for length 1 is done by taking the head of the list (all elements except the last one) and negating it (empty lists are falsy, non-empty lists are truthy).

Answer 7

JavaScript (ES6), 46 41 40 bytes

Takes input as a string. Returns RangeError if there are no unique digits.

s=>f=(i=9)=>s.split(i).length-2?f(--i):i

-7 bytes thanks to Rick Hitchcock

-1 byte thanks to Shaggy

Test cases

let f =

s=>g=(i=9)=>s.split(i).length-2?g(--i):i


console.log(f("12")())         // 2
console.log(f("0")())          // 0
console.log(f("485902")())     // 9
console.log(f("495902")())     // 5
//console.log(f("999999")())   // RangeError
console.log(f("999099")())     // 0
console.log(f("1948710498")()) // 7

Answer 8

Haskell, 37 bytes

f s=maximum[x|x<-s,[x]==filter(==x)s]

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How it works:

  [  |x<-s   ]          -- loop x through the input string s
    x                   -- and keep the x where
     [x]==filter(==x)s  -- all x extracted from s equal a singleton list [x]
maximum                 -- take the maximum of all the x

Answer 9

R, 41 bytes

function(x,y=table(x))max(names(y[y==1]))

An anonymous function that takes a list of digits, either as integers or single character strings. It precomputes y as an optional argument to avoid using curly braces for the function body. Returns the digit as a string. This takes a slightly different approach than the other R answer and ends up being the tiniest bit shorter! looks like my comment there was wrong after all...

table computes the occurrences of each element in the list, with names(table(x)) being the unique values in x (as strings). Since digits are fortunately ordered the same lexicographically as numerically, we can still use max.

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Answer 10

Python 3, 40 bytes

lambda i:max(x+9-9*i.count(x)for x in i)

Only works for lists of digits. The edge case '990' works fine :)

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Answer 11

Husk, 3 bytes

►≠O

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Explanation

►≠O
  O order the elements
►   max by
 ≠  inequality(selects least frequent elements)
    then returns the last of the least frequent elements

Answer 12

Brachylog, 8 bytes

ọtᵒtᵍhth

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Explanation

Example input: 495902

ọ          Occurences:    [[4,1],[9,2],[5,1],[0,1],[2,1]]
 tᵒ        Order by tail: [[0,1],[2,1],[4,1],[5,1],[9,2]]
   tᵍ      Group by tail: [[[0,1],[2,1],[4,1],[5,1]],[[9,2]]]
     h     Head:          [[0,1],[2,1],[4,1],[5,1]]
      t    Tail:          [5,1]
       h   Head:          5

Answer 13

APL (Dyalog Unicode), 10 chars = 19 bytes

Method: multiply elements that occur multiple times by zero, and then fine the highest element.

⌈/×∘(1=≢)⌸

⌸ for each unique element and its indices in the argument:

 × multiply the unique element

 ∘(…) with:

  1= the Boolean for whether one is equal to

  ≢ the tally of indices (how many times the unique element occurs)

⌈/ the max of that

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APL (Dyalog Classic), 15 bytes

⌈/×∘(1=≢)⎕U2338

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Identical to the above, but uses ⎕U2338 instead of ⌸.

Answer 14

Husk, 9 8 bytes

Thanks to Leo for suggesting a slightly neater solution at the same byte count.

▲‡ȯf=1`#

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Explanation

  ȯ       Compose the following thre functions into one binary function.
      `#  Count the occurrences of the right argument in the left.
    =1    Check equality with 1. This gives 1 (truthy) for values that 
          appear uniquely in the right-hand argument.
   f      Select the elements from the right argument, where the function
          in the left argument is truthy.
          Due to the composition and partial function application this
          means that the first argument of the resulting function actually
          curries `# and the second argument is passed as the second
          argument to f. So what we end up with is a function which selects
          the elements from the right argument that appear uniquely in
          the left argument.
 ‡        We call this function by giving it the input for both arguments.
          So we end up selecting unique digits from the input.
▲         Find the maximum.  

Answer 15

Mathematica, 41 bytes

(t=9;While[DigitCount[#][[t]]!=1,t--];t)&

thanks @Martin Ender

here is Martin's approach on my answer

Mathematica, 35 bytes

9//.d_/;DigitCount[#][[d]]!=1:>d-1&

Answer 16

R, 45 43 bytes

function(x)max(setdiff(x,x[duplicated(x)]))

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Takes input as a vector of integers. Finds the duplicated elements, removes them, and takes the maximum. (Returns -Inf with a warning if there is no unique maximum.)

Edited into an anonymous function per comment

Answer 17

Python 2, 39 bytes

lambda l:max(1/l.count(n)*n for n in l)

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Answer 18

Ruby, 42 bytes

->x{(?0..?9).select{|r|x.count(r)==1}[-1]}

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Answer 19

Bash + coreutils, 30 28 bytes

-2 bytes thanks to Digital Trauma

fold -1|sort|uniq -u|tail -1

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---

Bash + coreutils, 20 bytes

sort|uniq -u|tail -1

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If input is given as a list of digits, one per line, we can skip the fold stage. That feels like cheating though.

Answer 20

C# (.NET Core), 27 97 86 58 57 75 bytes

using System.Linq;

n=>n.GroupBy(i=>i).Where(i=>i.Count()<2).Max(i=>i.Key)-48

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Thanks @CarlosAlejo

Answer 21

<>< (Fish), 93 bytes

0l1-l2-$:?v~~o;         
v&]{[-3lr <   |.!01~$r&\
>r$1-:?!v$rl3-[{]:&:&=?^
@&::&:$~/|.!01 $}&~$?(

Try it in my interpreter

Found a fun bug in my interpreter testing this

Expects input pre-pushed to the stack

Explanation

Top row: Setup. Push 0, the current maximum, l-1, the total number of iterations of the outer loop, and l-2, the total number of iterations in the inner loop. Initially both are the same.

Second row left: Shift left 1, then move the top item to the register. This is the candidate largest member.

Third row: Compare the register for each digit. If none match go down. Otherwise go up to second row right.

Second row right: Reset, then try another digit. This happens if the digit is equal to another.

Last row: We have found a unique digit! Now we check if it's better than the current max value, if so replace it. Then reset and jump to the start.

Answer 22

JavaScript (ES6), 52 50 bytes

Takes input as a list of digits. Returns 0 if there are no unique digits.

s=>s.reduce((m,c)=>m>c|s.filter(x=>x==c)[1]?m:c,0)

Test cases

let f =

s=>s.reduce((m,c)=>m>c|s.filter(x=>x==c)[1]?m:c,0)

console.log(f([1,2]))                 // 2
console.log(f([0]))                   // 0
console.log(f([4,8,5,9,0,2]))         // 9
console.log(f([4,9,5,9,0,2]))         // 5
console.log(f([9,9,9,9,9,9]))         // (0)
console.log(f([9,9,9,0,9,9]))         // 0
console.log(f([1,9,4,8,7,1,0,4,9,8])) // 7

Answer 23

Java (OpenJDK 8), 89 85 79 bytes

a->{int i=10,x[]=new int[i];for(int d:a)x[d]++;for(;i-->0&&x[i]!=1;);return i;}

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-6 bytes thanks to @KevinCruijssen's insight!

Answer 24

APL (Dyalog), 14 bytes

-2 thanks toTwiNight.

⌈/⊢×1=(+/∘.=⍨)

⌈/ the largest of

⊢ the arguments

× multiplied by

1=(…) the Boolean for each where one equals

 +/ the row sums of

 ∘.=⍨ their equality table

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Answer 25

PHP, 40 bytes

<?=array_flip(count_chars($argn))[1]-48;

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PHP, 42 bytes

<?=chr(array_flip(count_chars($argn))[1]);

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Answer 26

Java (JDK), 67 bytes

s->{int i=9;for(s=" "+s+" ";s.split(i+"").length!=2;i--);return i;}

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Answer 27

J, 16 12 bytes

0{\:~-.}./.~

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Answer 28

Japt -h, 12 11 10 4 bytes

I/O as a digit array.

ü l1

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Answer 29

APL (NARS2000) 16 chars, 23(?) bytes

{↑⌽⍸1=+⌿⍵∘.=⍳10}

(input as array of digits, e.g. {↑⌽⍸1=+⌿⍵∘.=⍳10}4 9 5 9 0 2)

Answer 30

Desmos, 113 bytes

U(l)=max([min([\left\{l[n]=l[i]:-1,l[n]\right\} for i=join([0...n-1],[n+1...l.length+1])]) for n=[1...l.length]])

Takes input as a list.

Try it out and see versions that make slightly more sense