In Search of a Soulmate

Question

Given a nonempty finite list of integers, output a truthy value if there are exactly two equal entries and all other entries are distinct, and a falsey value otherwise.

Examples

truthy:
[1,1]
[1,2,1]
[1,6,3,4,4,7,9]

falsey:
[0]
[1,1,1]
[1,1,1,2]
[1,1,2,2]
[2,1,2,1,2]
[1,2,3,4,5]

Answer 1

Python 3, 30 28 bytes

lambda m:len({*m})+1==len(m)

Try it online!

{*m} casts the list to a set object, an unordered list of items without duplicates. Doing this will always decrease the length of the list by the number of duplicates in it. By computing how much the length has changed, we can easily tell if the list had a single duplicate and return the result of the test.

^{-2 bytes thanks to ovs.}

Answer 2

Husk, 4 bytes

εṠ-u

Try it online!

Explanation

εṠ-u  Implicit input.
   u  Unique elements.
 Ṡ-   Delete them from input, counting multiplicities.
ε     Is the result a singleton list?

Answer 3

MATL, 7, 6 bytes

&=sp4=

Try it online!

One byte saved thanks to @Guiseppe!

Explanation:

&=  % Table of pair-wise equality comparisons
    %
    % [1 0 0 0 0 0 0
    %  0 1 0 0 0 0 0
    %  0 0 1 0 0 0 0
    %  0 0 0 1 1 0 0
    %  0 0 0 1 1 0 0
    %  0 0 0 0 0 1 0
    %  0 0 0 0 0 0 1]
    %
s   % Sum each Column. Stack:
    %
    % [1 1 1 2 2 1 1]
    %
p   % Product of the array. Stack:
    %
    % [4]
    %
4=  % Compare the stack to '4'

Answer 4

Haskell, 34 bytes

f x=[1|a<-x,b<-x,a==b]==1:1:(1<$x)

Try it online! Based on H.PWiz' answer.

Answer 5

Jelly, 8 5 bytes

QL‘=L

Try it online!

Explanation

QL‘=L  - Main link, argument L (a list)   e.g [1,6,3,4,4,7,9]
Q      - Deduplicated elements                [1,6,3,4,7,9]
 L     - Length                               6
  ‘    - Increment                            7
    L  - Length of the input                  7 ([1,6,3,4,4,7,9])
   =   - Are they equal?                      1

If the output values can be any consistent values, then QL_L works, which outputs -1 for truthy and any other non-positive number for falsey (thanks @JonathanAllan)

Answer 6

JavaScript (ES6), 30 bytes

a=>new Set(a).size==a.length-1

Try it online

Answer 7

Retina, 15 12 11 bytes

Thanks to Neil for saving 1 byte.

D`
Mm2`^$
1

Try it online!

Input is linefeed-separated. (The test suite uses comma-separation for convenience.)

Explanation

D`

Deduplicate the lines in the input, which removes any integer that has appeared before (but leaves the surrounding linefeed(s)).

Mm2`^$

Count the number of empty lines, which is equal to the number of duplicates we removed, but only consider the first two matches. So the output will only be 0 (no duplicates), 1 (one duplicate), 2 (two or more duplicates).

1

Make sure that exactly one duplicate was removed.

Answer 8

J, 7 6 bytes

=&#0,=

= check every element for equality with every unique element, creates a matrix with m rows for m unique elements.
0, add an empty row on top.
=&# does the number of rows equal the length of input?

Try it online!

Answer 9

Octave, 25 bytes

This is not using a group or unique approach as many of the other answers, but rather the "cartesian product" of all possible comparisions.

@(x)nnz(triu(x==x',1))==1

Explanation

             x==x'        %create a matrix where the entry at (i,j) compares whether x(i) == x(ju)
        triu(x==x',1)     %only consider the strict upper triangular matrix
    nnz(triu(x==x',1))    %count the number of nonzero entries
@(x)nnz(triu(x==x',1))==1 %check whether this number is actually 1

Try it online!

And because no program would be complete without a convolution (thanks @LuisMendo for fixing a mistake):

Octave, 40 bytes

@(x)nnz(~conv(sort(x),-1:2:1,'same'))==1

Try it online!

Answer 10

Pushy, 8 bytes

Simple implementation of checking whether len(set(list)) == len(list)-1:

LtvuL^=#

Explanation:

       \ Implicit: Put all input on stack
Ltv    \ Get the stack length - 1, save in auxiliary stack
u      \ Remove non-unique elements
L      \ Get the new length
^=     \ Compare with the previously saved length
#      \ Print result

This works as the length will only decrease by 1 if there was only exactly 1 non-distinct integer in the initial list.

Try it online!

Answer 11

PowerShell, 40 37 bytes

($args|sort -u).count-eq$args.count-1

Try it online!

The Sort-Object command (alias sort) with the -unique flag pulls out only the unique components of the input. For example, for input @(1,3,3,2), this will result in @(1,2,3).

Thus, we just need to make sure that the .count of this object (i.e., how many elements it has) is -equal to the .count of our input array -1 (i.e., we have exactly one duplicate entry).

Saved 3 bytes thanks to Sinusoid.
Fixed bug thanks to TessellatingHeckler.

Answer 12

R, 32 31 bytes

-1 byte thanks to @JarkoDubbeldam

cat(sum(duplicated(scan()))==1)

Try it online!

Reads from stdin, writes to stdout.

duplicated iterates through the list, replacing the values of l with TRUE if that value occurs earlier in the list, and FALSE otherwise. If there's a unique pair of soulmates, there should be exactly one TRUE value, so the sum should be 1.

Answer 13

05AB1E, 4 bytes

{¥_O

Try it online!

Outputs 1 as truthy, any other non-negative integer as falsy. In 05AB1E, 1 is the only truthy number (thanks @Emigna for the insight!).

Explanation

{       Implicit input. Sort
 ¥      Consecutive differences
  _     Boolean negate
   O    Sum. Implicit display

Answer 14

Ruby, 32 bytes

->(s){s.uniq.length==s.length-1}

Answer 15

C# (.NET Core), 35 + 18 bytes

+18 for using System.Linq.

n=>n.Distinct().Count()==n.Length-1

Try it online!

67 byte alternative without Linq:

n=>new System.Collections.Generic.HashSet<int>(n).Count==n.Length-1

Try it online!

Answer 16

Excel, 42 bytes

Danish language version

=TÆLV(A:A)=SUM(--(FREKVENS(A:A,A:A)>0))+1

Assumes each integer from the list in separate cell in column A.
If we were allowed for inconsistent falsey values, we could save 3 bytes:

=TÆLV(A:A)+SUM(-(FREKVENS(A:A,A:A)>0))

English language version (44 bytes)

=COUNTA(A:A)=SUM(--(FREQUENCY(A:A,A:A)>0))+1

Answer 17

05AB1E, 6 5 bytes

{¥>ΘO

Try it online!

{¥>ΘO   # example input:               [1, 6, 3, 4, 4, 7, 9]
{       # sort                      -> [1, 3, 4, 4, 6, 7, 9]
 ¥      # get deltas                -> [  2, 1, 0, 2, 1, 2 ]
  >     # increment                 -> [  3, 2, 1, 3, 2, 3 ]
   Θ    # truthify (only 1 gives 1) -> [  0, 0, 1, 0, 0, 0 ]
    O   # sum                       -> 1

1 being the only truthy value in 05AB1E, we can stop here. (Thanks @Emigna for pointing that out.)

To get only two distinct values, we can optionally add:

     Θ  # equals 1?                 -> 1

Answer 18

Perl 5, 36 + 1 (-a) = 37 bytes

map$k{$_}++,@F;@a=keys%k;say@a+1==@F

Try it online!

Answer 19

Haskell, 37 bytes

f x=sum[1|a<-x,b<-x,a==b]==2+length x

Try it online!

Answer 20

Octave / MATLAB (with Statistics package / toolbox), 21 bytes

@(x)nnz(~pdist(x))==1

Anonymous function. Input is a column vector. Output is true (displayed as 1) or false (displayed as0).

Try it online!

Explanation

pdist(x) computes a vector of Euclidean distances between all pairs of rows from x. It considers each pair only once (order of the two rows doesn't matter), and doesn't consider pairs formed by the same row twice.

In our case x is a column vector, so Euclidean distance between two rows is just absolute difference between the two numbers.

~ is logical (Boolean) negation, nnz is number of nonzeros, and ==1 compares to 1. So the result is true if and only if there is only one pair that gives zero distance.

Answer 21

Julia, 39 26 bytes

!a=sum(a.==a')==endof(a)+2

Explanation

The code generates a 2-dimensional table of booleans, which is then collected using the sum function, counting the number of same-element pairs in the cartesian square of A. Then this is compared to the length of the string plus two, and the quantities are equal only when there is exactly one repeat character.

This code redefines the NOT operator.

Answer 22

Jq 1.5, 53 25 bytes

length-(unique|length)==1

Inspired by Riley's answer and much shorter then my original solution.

Try it online!

Answer 23

Pyth, 6 bytes

qtlQl{

Verify all the test cases.

• l{ - Gets the number of unique elements.

• tlQ - Gets the length of the input list, decremented.

• q - Checks equality.

7 bytes

q1l.-Q{

Verify all the test cases

Answer 24

Octave, 23 26 bytes

@(x)prod(sum(x==x'))==4

Try it online!

The x==x' part was inspired by flawr's answer. This is longer than Luis' answer, but it doesn't use any toolboxes.

Explanation:

This is an anonymous function that takes a vector x as input, and compares it to itself transposed. This will give a matrix where all diagonal elements are 1, and any off diagonal elements signals that there are duplicates elements.

The sum along any given column shows how many duplicates there are of that number. We want two of the numbers to have a duplicate, so we two values equal to two, and the rest unequal to two.

If we take the product of this matrix, we'll get 4 if there are only two equal elements (2*2*1*1*1*1*...), and something other than 4 if there are no duplicates, or more than two.

Answer 25

PHP, 46 bytes

<?=count(array_unique($argv))==count($argv)-1;

Counts the number of entries in $argv and compares it to the number of unique entries. If the former is higher than the latter by 1 then truthy, else falsey.

Try it on eval.in!

Answer 26

APL (Dyalog Unicode), 7 bytes^SBCS

1=≢-≢∘∪

Try it online!

Explanation:

1=≢-≢∘∪  ⍝ Monadic function train
    ≢∘∪  ⍝ Generate a list of unique items in the input,
         ⍝ and return the length of that list
  ≢-     ⍝ Take the length of the input and subtract the above
1=       ⍝ If the difference is 1, true, otherwise false

Answer 27

Jelly, 10 bytes

ċ@€`QṢ⁼1,2

Try it online!

a longer but different approach

Answer 28

Japt, 7 bytes

â ʶUÊÉ

Try it

---

Explanation

Remove duplicates (â), get length (Ê) and compare equality (¶) with the length (Ê) of the input (U) minus 1 (É).

Answer 29

Haskell, 37 bytes

f x=sum[1|0<-(-)<$>x<*>x]==2+length x

Try it online!

Answer 30

05AB1E, 5 bytes

gIÙg-

Try it online!

g     # Get number of elements in input
 IÙg  # Get number of unique elements in input
    - # Subtract

In 05AB1E 1 is the only truthy value, so for a truthy result there must be exactly 1 duplicate element removed by the uniquify.

Answer 31

Bash + coreutils, 36 bytes

sort|uniq -dc|grep -Pqz '^ *2 .*\n$'

Output is via exit code, where 0 is success (truthy) and 1 is failure (falsy).

Try it online!

Answer 32

C (GCC), 80 78 bytes

i,j,t;f(x,l)int*x;{for(i=t=0;i<l;++i)for(j=0;j<i;)t+=x[i]==x[j++];return!~-t;}

-1 thanks to Jonathan Frech

-1 thanks to Kevin Cruijssen

Try it Online!

f is a function that takes in an int* pointing to the list, and an int that is the length of the list, and returns 1 if there are exactly two equal entries and all other entries are distinct, and 0 value otherwise.

The function checks all pairs of numbers in the list, counting the number of pairs, and returns whether the number of pairs is 1.

Answer 33

Java 8, 46 44 bytes

l->l.stream().distinct().count()==l.size()-1

-2 bytes thanks to @Nevay. (Old answer: l->new java.util.HashSet(l).size()==l.size()-1)

Explanation:

Try it here.

l->                // Method with List parameter and boolean return-type
  l.stream()       //  Stream over the List
   .distinct()     //  ignoring all duplicated items
   .count()        //  and get the total amount of non-duplicated items in the List
     ==            //   And check if it's size is equals to
       l.size()-1  //   the size of the input-list - 1
                   // End of method (implicit / single-line return-statement)

Answer 34

Bash, 36, 35 bytes

for k;{ H[$k]=;};((${#H[@]}+1==$#))

TIO exit status 0: true, 1: false, ((..)) can be changed to echo $((..)), to see boolean value (1:true, 0:false)

Answer 35

Clojure, 31 bytes

#(=(count(set %))(-(count %)1))

Try it online!

Does the same as LyricLy's answer

Answer 36

Japt, 7 bytes

ÊɶUâ Ê
      Ê // Return whether the number of
   Uâ   // unique items in the input
  ¶     // is equal to
ÊÉ      // the input's length minus one.

Try it online!

Answer 37

Brachylog, 7 bytes

dl.&l-₁

Try it online!

Truthy/falsy input is achieved through predicate success/failure, as if this predicate is run as the entire program on a single input, it will print true. if it succeeds and false. if it fails. The header on TIO is there so you can run all of the cases at once.

dl         The length of the input with duplicates removed
  .        is the output variable,
   &       and
    l-₁    so is the length of the input minus 1.

Answer 38

Regex (ECMAScript), 69 bytes

The input is in the form of a comma-delimited list of nonnegative integers in decimal.

^(?=.*(\b\w+\b).*\b\1\b)(?!(.*\b\1\b){3}|.*\b(?!\1\b)(\w+\b).*\b\3\b)

Try it online!

^
(?=
    .*(\b\w+\b)          # \1 = an element that occurs at least twice
    .*\b\1\b             # locate the second occurrence of \1
)
(?!
    (.*\b\1\b){3}        # Assert that \1 does not occur 3 or more times
|
    .*\b(?!\1\b)(\w+\b)  # \3 = any element that's different from \1
    .*\b\3\b             # Assert that \3 does not occur again
)

This can be trivially modified to work on positive integers in unary instead of nonnegative integers in decimal, in -2 bytes (67) by changing \w+ to x+. Making it handle nonnegative integers in unary would be a tad more involved, due to \b not matching between two commas.

Answer 39

Mathematica, 26 bytes

(l=Length)@Union@#+1==l@#&

Try it online!

Answer 40

Convex, 7 bytes

_Å,)\,=

Try it online!

Answer 41

Batch, 109 bytes

@set/ap=1,s=0
@for %%x in (%*)do @(for %%y in (%*)do @set/a"s+=!(%%x-%%y)")&set/ap*=s,s=0
@if %p%==4 echo 1

Port of @DJMcMayhem's answer.

Answer 42

Perl 5, 25+1 (-p)=26 bytes

$H{$_}=1}{$\=$.-1==keys%H

TIO

Answer 43

Batch, 131

set b=,%*,
goto j
:l
call set f=%%b:,%1,=,#,%%
if "%b%"=="%f%" exit 1
set b=%f%
goto:eof
:j
FOR %%A IN (%b%) DO call :l %%A

The %errorlevel% can be checked for the result. it works by tokenising the input, FOR %%A IN (%b%) DO call :l %%A every instance of that token + to commas is then replaced in a copy of the input with a ",#," call set f=%%b:,%1,=,#,%% the resulting string is then compared to the copy of the previous string. If there is no change in the string, it is because that number has already occured.

c:\>batfile.bat 1,2,3

Answer 44

Perl 6, 12 bytes

{.Set+1==$_}

Try it online!

Convert to Set, test if the number of elements is less by one.

Answer 45

R, 40 bytes

sum(outer(x<-scan(),x,"=="))==2+sum(x|1)

Try it online!

Approach is different from the other R solution - and not as golfy...

Answer 46

12-basic, 31 bytes

FUNC S(L)?LEN(L|L)==LEN(L)-1END

The | (union) operator returns an array containing elements that are in both of the input arrays. As a side effect, it removes duplicates.

Answer 47

Pari/GP, 16 bytes

a->#a==#Set(a)+1

Try it online!

Answer 48

R, 38 bytes

Not quite as golfed as another R answer, but a different approach.

length(a<-scan())-length(unique(a))==1

Try it online!

Answer 49

F#, 49 bytes

let s c=Seq.distinct c|>Seq.length=Seq.length c-1

Try it online!

Seq.distinct creates a sequence of all the unique elements in the collection. If the number of distinct elements is one less than the original collection, there are soulmates.

Answer 50

Javascript,96 bytes

for(i=0;i<a.length;i++)for(j=0;j<a.length;j++)(a[i]==a[j]&& i!=j)?count++:0 console.log(count==2)

And a better readable format

for(i=0;i<a.length;i++){
  for(j=0;j<a.length;j++){
   (a[i]==a[j]&& i!=j)?count++:0
  }
}
 console.log(count==2)