The shortest way to find one unique value when all other values are the same

Question

In this question at Code Review they tried to find the fastest way to return the unique element in an array where all the elements are the same except one. But what is the shortest code that accomplish the same thing?

Goal

Find the unique element in an array and return it.

Rules

• The input array will contain only integer, strictly positive numbers, so you can use 0 as the end of the input if your language needs it.
• The size of the array will be at least 3 and will have a finite size. You can limit the size of the array to any limit your language has.
• Every element in the array will be the same, except for one which will be different.
• You must output the value (not the position) of the unique element in any standard format. You can output leading or trailing spaces or newlines.
• You can take the input array in any accepted format.

Examples

Input                   Output
------------------------------
[ 1, 1, 1, 2, 1, 1 ]      2
[ 3, 5, 5, 5, 5 ]         3
[ 9, 2, 9, 9, 9, 9, 9 ]   2
[ 4, 4, 4, 6 ]            6
[ 5, 8, 8 ]               5
[ 8, 5, 8 ]               5
[ 8, 8, 5 ]               5

Winner

This is code-golf, so may the shortest code for each language win!

Answer 1

Python 3, 27 bytes

lambda x:min(x,key=x.count)

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Answer 2

R, 24 bytes

a=scan();a[a!=median(a)]

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Answer 3

JavaScript (ES6),  32  27 bytes

Saved 5 bytes thanks to @xnor

a=>a.sort()[0]+a.pop()-a[1]

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How?

a.sort() sorts the input array in lexicographical order. We know for sure that this is going to put the unique element either at the first or the last position, but we don't know which one:

[ x, ..., y, ..., x ].sort() -> [ y, x, ..., x ] or [ x, x, ..., y ]

Either way, the sum of the first and the last elements minus the 2nd one gives the expected result:

[ y, x, ..., x ] -> y + x - x = y
[ x, x, ..., y ] -> x + y - x = y

We could also XOR all of them:

a=>a.sort()[0]^a.pop()^a[1]

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Answer 4

x86-16 machine code, 14 bytes

Binary:

00000000: 498d 7c01 f3a6 e305 4ea6 7401 4ec3       I.|.....N.t.N.

Listing:

49          DEC  CX             ; only do length-1 compares 
8D 7C 01    LEA  DI, [SI+1]     ; DI pointer to next value 
F3 A6       REPE CMPSB          ; while( [SI++] == [DI++] );
E3 05       JCXZ DONE           ; if end of array, result is second value
4E          DEC  SI             ; SI back to first value
A6          CMPSB               ; [SI++] == [DI++]? 
74 01       JE   DONE           ; if so, result is second value
4E          DEC  SI             ; otherwise, result is first value 
        DONE: 
C3          RET                 ; return to caller 

Callable function, input array in [SI], length in CX. Result in [SI].

Explanation:

Loop through the array until two different adjacent values are found. Compare the first to the third value. If they are the same, the "odd value out" must be the second, otherwise it is the first.

Example:

Input [ 1, 1, 1, 2, 1, 1 ], reduce until different adjacent values are found a = [ 1, 2, 1, 1 ]. If a[0] == a[2] then result is a[1], otherwise result is a[0].

Tests using DOS DEBUG:

Answer 5

APL (Dyalog Extended), 5 bytes

⍸1=¯⍸

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Regular Dyalog APL 18.0 has ⍸⍣¯1, but it doesn't work here because it requires the input array to be sorted, unlike Extended's ¯⍸ which allows unsorted input arrays.

How it works

⍸1=¯⍸  ⍝ Input: a vector N of positive integers
       ⍝ (Example: 4 4 6 4)
   ¯⍸  ⍝ Whence; generate a vector V where V[i] is the count of i in N
       ⍝ (¯⍸ 4 4 6 4 → 0 0 0 3 0 1)
 1=    ⍝ Keep 1s intact and change anything else to zero (V1)
       ⍝ (1= 0 0 0 3 0 1 → 0 0 0 0 0 1)
⍸      ⍝ Where; generate a vector W from V1, where i appears V1[i] times in W
       ⍝ (⍸ 0 0 0 0 0 1 → 6)

Answer 6

Haskell, 32 bytes

f(x:y)|[e]<-filter(/=x)y=e|1<3=x

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Explanation

The code, expanded with variables renamed to be more descriptive.

f (first:rest)
 | [unique] <- filter (/=first) rest = unique
 | 1 < 3 = first

(first:rest) is a pattern match on a list that destructures it into its first element (first) and the list without the first element (rest).

Each line with a | at the front is a case in the function (known as "guards" in Haskell). The syntax looks like functionName args | condition1 = result1 | condition2 = result2 .... There are two cases:

1. [unique] <- filter (/=first) rest. This asserts that filter (/=first) rest produces a list containing only one element, which we name unique. filter (/=first) rest filters out all elements in rest not equal to first. If we are in this case, then we know that unique is the unique element, and we return it.
2. 1 < 3. This asserts that 1 is less than 3. Since it's always true, this is a "fallthrough" case. If we reach it, we know that there are at least 2 elements not equal to the first element, so we return first.

Answer 7

05AB1E, 2 bytes

ʒ¢

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Since 1 is the only truthy integer in 05AB1E, we can just filter (ʒ) on the ¢ount.

There is also the Counter-Mode builtin, which returns the least frequent element in a list at the same bytecount:

.m

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Answer 8

Bash + coreutils, 12 bytes

sort|uniq -u

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Answer 9

Jelly, 3 bytes

ḟÆṃ

Explanation: ḟ (probably) removes all elements that are not the most common element (returned by Æṃ). I don't know why isn't the result a single-element list (perhaps it's a feature I didn't know about?), but that makes this even better.

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Answer 10

J, ¹⁰ 8 bytes

1#.|//.~

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How it works

 1#.|//.~  input:                  1 1 1 2 1
      /.~  group by itself:        1 1 1 1
                                   2
    |/     insert | (remainder) into both groups:        
                   1 | 1 | 1 | 1 = 0
                   2             = 2
 1#.       sum:                    2

Answer 11

Octave, 17 bytes

@(x)x(x!=mode(x))

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2 bytes golfed thanks to Luis Mendo!

Answer 12

Octave / MATLAB, 19 bytes

@(x)x(sum(x==x')<2)

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How it works

@(x)                  % Define a function of x
          x==x'       % Square matrix of all pairwise comparisons (non-complex x)
      sum(     )      % Sum of each column
                <2    % Less than 2? This will be true for only one column
    x(            )   % Take that entry from x

Answer 13

Wolfram Language (Mathematica), 21 19 17 bytes

f[c=a_...,b_,c]=b

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Answer 14

R, 25 bytes

a=sort(scan());a[a!=a[2]]

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Sort the input, giving a. Now a[2] is one of the repeated values. Keep only the element not equal to a[2].

This ends up 4 bytes shorter than my best shot with a contingency table:

names(which(table(scan())<2))

Answer 15

Pyth, 4 bytes

-1 byte thanks to @FryAmTheEggman

ho/Q

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Explanation

 o   : Order implicit input
  /Q : by count of the element
h    : then take the first element

Answer 16

Ruby, 18 bytes

->a{a-[a.sort[1]]}

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The unique element will always be the largest or smallest, so remove all copies of the second element.

Answer 17

RAD, 13 bytes

⍵[1⍳⍨+/⍵∘=¨⍵]

The lack of a need for {} in this language really helps.

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Explanation

⍵[1⍳⍨+/⍵∘=¨⍵]
     +/⍵∘=¨⍵  count of each element's # of occurrences
  1⍳⍨         first occurrence of a 1
⍵[          ] the argument at that index

Answer 18

K (ngn/k), 8 6 bytes

Solution:

*<#'=:

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Explanation:

*<#'=: / the solution
    =: / group the input
  #'   / count length of each
 <     / sort ascending
*      / take the first

Extra:

• -2 bytes thanks to ngn by dropping the lambda

Answer 19

Wolfram Language (Mathematica), 19 bytes

Tr[#/.Median@#->0]&

Inspired by Kirill L.'s R submission

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Finds the median, replaces its occurrences in the list with 0, and sums the list.

Answer 20

T-SQL, 40 bytes

Input is a table variable

DECLARE @ table(v int)
INSERT @ values(1),(1),(2),(1)

SELECT*FROM @ 
GROUP BY v
HAVING SUM(1)=1

Another variation

T-SQL, 54 bytes

DECLARE @ table(v real)
INSERT @ values(1),(1),(2),(1)

SELECT iif(max(v)+min(v)<avg(v)*2,min(v),max(v))FROM @

Answer 21

Brachylog, 4 bytes

oḅ∋≠

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Takes a list and returns a singleton list.

Explanation

oḅ∋≠  Input is a list, say [4,4,3,4].
o     Sort: [3,4,4,4]
 ḅ    Blocks of equal elements: [[3],[4,4,4]]
  ∋   Pick a block: [3]
   ≠  This block must have distinct elements (in this case, must have just one).
      Output it implicitly.

Answer 22

Gaia, 2 bytes

C⌡

Not a built-in!

C is "count occurrences in list" and ⌡ is "find the smallest by".

(the footer is "evaluate" and "call the function above", Gaia doesn't seem to evaluate input by default)

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Answer 23

PowerShell, 23 bytes

$args-ne($args|sort)[1]

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Takes input by splatting. We apply a filter to the original array getting rid of the dupes by sorting it and taking the 2nd element.

Answer 24

Wolfram Language (Mathematica), 27 bytes

Keys@Select[Counts@#,#<2&]&

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Answer 25

Haskell, 30 bytes

f(x:y)|elem x y=f$y++[x]|1>0=x

Try it online! Footer stolen from coles answer.

If the first element is not in the remaining list, return it, otherwise rotate it to the end and try again.

Answer 26

Java (JDK), 62 bytes

int f(int[] a){Arrays.sort(a);return a[0]+a[a.length-1]-a[1];}

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Thanks to JollyJoker for correct way.

Original completely own failing dumb#ss solution:

int f(int[] a){Arrays.sort(a);return a[0];}

Answer 27

C (gcc), 50 42 bytes

f(int*b,int*e){b=(*b^*e--?:f(b,e)^*e)^*e;}

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Answer 28

Excel, 14 bytes

Put the numbers in column A, and enter this formula in another column

=UNIQUE(A:A,,1

Answer 29

C (gcc), 39 bytes

f(int*a){a=*a-*++a?a[*a--!=a[2]]:f(a);}

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f(int*a){      // function taking a 0-terminated array pointer.

 a=            // return using eax register trick.
 *a-*++a?      // compare current item with next item :
 a[*a--!=a[2]] // if different we move the pointer by the result of comparing the next item,
:f(a);         // if not we recursively call on the incremented pointer.
}

Answer 30

C GCC, 40 bytes

f(int*a){a=*a^*++a?a[-(*a==a[1])]:f(a);}

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Thanks to xibu for -5 bytes!

Thanks to ceilingcat for -2 bytes!