Equal numbers in sub-array

Question

Given an array of numbers with length >=3 and length % 3 == 0

[1, 2, 3, 4, ...]

You will split it in sub-arrays of length 3

[[1, 2, 3], [4, 5, ...], [...

And return an array with

• [0] => The amount of cases in the sub-array where all numbers are equal
• [1] => In case all numbers in sub-array are not equal, the amount of cases in the sub-array where only 2 numbers are equal

Example and test cases:

• Input: [2, 4, 2, 5, 5, 5, 4, 2, 1, 3, 3, 1] output [1, 2]

This is because

[[2, 4, 2], [5, 5, 5], [4, 2, 1], [3, 3, 1]]
  ^     ^    ^  ^  ^               ^  ^ 
   equal    all equal              equal   

so 2 equal and 1 all equal

• [3,5,6,5,5,7,6,6,8,7,7,7,3,4,2,4,4,3] => [1, 3]
• [3,3,3,4,4,4,5,5,5,6,6,6,5,4,3] => [4, 0]
• [3,4,5,6,7,8,9,8,7,6,5,4,3,2,1] => [0, 0]

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This is code-golf, so the shortest answer in bytes win.

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PD: Apologies for my English.

Answer 1

Octave, 60 52 50 bytes

@(x)sum(sum(~diff(sort(reshape(x,3,[]))))'==[2 1])

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Saved 8 bytes thanks to Luis!

Explanation:

Reshapes the input into a matrix with 3 rows, and the appropriate amount of columns. It then sorts each of the columns, and calculates the difference between the elements on different rows. This gives a matrix with two rows, where identical numbers will have a zero, and different numbers will have a positive number. This is negated, so that all equal elements are 1, and all unequal are 0. We then sum each of those columns, giving us one of the three alternatives: 0 = All elements are unequal, 1 = Two elements are equal and 2 = All elements are equal. We then check how many are >1, and how many are exactly ==1.

Answer 2

Jelly,  9  8 bytes

-1 thanks to Dennis (use a new alias for L€, Ẉ)

Q3ÐƤẈċⱮ2

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Answer 3

05AB1E, 10 bytes

3ôεÙg}12S¢

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Explanation

3ô          # split input into groups of 3
  ε  }      # for each triple
   Ù        # remove duplicates
    g       # and get the length
      12S¢  # count the number of 1s and 2s in the result

Answer 4

JavaScript (ES6), 70 bytes

f=([a,b,c,...d],t=p=0)=>1/a?f(d,t+!(a-b&&a-c?b-c||++p:b-c&&++p)):[t,p]

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How?

We recursively extract each triplet [a,b,c] from the input array and update two counters t (three-of-a-kind) and p (pair), using the following formula:

t =
t + !(a - b && a - c ? b - c || ++p : b - c && ++p)

There are 5 possible cases which are detailed below, from 'all equal' to 'all distinct'.

a b c | a-b && a-c | b-c | b-c || ++p | b-c && ++p | t +=
------+------------+-----+------------+------------+------------
4 4 4 | false      | 0   | n/a        | 0          | !0    --> 1
4 4 5 | false      | ≠0  | n/a        | ++p        | !++p  --> 0
4 5 4 | false      | ≠0  | n/a        | ++p        | !++p  --> 0
5 4 4 | true       | 0   | ++p        | n/a        | !++p  --> 0
4 5 6 | true       | ≠0  | ≠0         | n/a        | !(≠0) --> 0

Answer 5

Pyth, 13 14 12 11 bytes

/Lml{kcQ3S2

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Explanation

/Lml{kcQ3S2
      cQ3        Split the input into groups of 3.
  ml{k           Deduplicate and get the length of each.
/L               Count the number...
         S2      ... of 1s and 2s.

Answer 6

oK, 17 16 bytes

+/(1 2=#=:)'0N3#

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            0N3# /reshape into groups of 3 (see ngn's comment)
  (       )'     /for each group:
        =:       /    make a map from number -> indices
       #         /    count number of keys/values
   1 2=          /    check if the count is equal to 1 or 2 
+/               /sum together the columns

For k, the 17 byte version is: +/(1 2=#=:)'0N 3#.

Answer 7

R, 70 bytes

function(v,x=lengths(by(v,seq(0,a=v)%/%3,table)))c(sum(x<2),sum(x==2))

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Previous versions :

R, 82 bytes

function(v,a=!1:2){for(i in lengths(by(v,seq(0,a=v)%/%3,table)))a[i]=a[i]+1;a[-3]}

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R, 93 bytes

function(v,a=table(lengths(by(v,0:(length(v)-1)%/%3,unique)))[c('1','2')])`[<-`(a,is.na(a),0)

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Answer 8

Java (JDK 10), 116 bytes

l->{int r[]={0,0,0},i=0,a,b,c;for(;i<l.length;b=l[i++],c=l[i++],r[a==b?b==c?0:1:b==c|a==c?1:2]++)a=l[i++];return r;}

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Note: returns a 3-elements array with [0] and [1] returning th appropriate values, and [2] returning a dummy value (the number of 3-lists without any elements in common) . This is totally valid according to the current rules.

Answer 9

PowerShell, 106 bytes

param($a)for(;$a){$x,$y,$z,$a=$a;if($x-eq$y-and$y-eq$z){$i++}else{$j+=$x-eq$y-or$y-eq$z-or$z-eq$x}}+$i,+$j

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Exactly what it says on the tin. Loops over input $a. Each iteration, peels off $x,$y,$z as the next three elements. Tests if they're all equal and if so, increments $i. Else, increments $j if at least one pair is equal. Once the loop is complete, output $i and $j as integers.

So ... many ... dollars ...

Answer 10

Retina 0.8.2, 68 bytes

(.+)¶(.+)¶(.+)
;$1;$2;$3;$1;
%M`(;\d+)(?=\1;)
s`((1)|(3)|.)+
$#3 $#2

Try it online! Link includes test cases with header to convert to desired format of one value per line. Explanation:

(.+)¶(.+)¶(.+)
;$1;$2;$3;$1;

Collect three values onto each line with separators and duplicate the first one at the end.

%M`(;\d+)(?=\1;)

Count the number of pairs of duplicates.

s`((1)|(3)|.)+
$#3 $#2

Count the number of 3s and 1s.

Answer 11

Stax, 8 bytes

íUÖ←#"ë╕

Run and debug it

Answer 12

Common Lisp, 113 bytes

(lambda(l &aux(a 0)(b 0))(loop for(x y z)on l by #'cdddr do(if(= x y z)(incf a)(if(/= x y z)()(incf b))))`(,a,b))

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Used the fact that in Common Lisp (= x y z) gives true if all the three elements are equal, and (/= x y z) gives true if no pair of numbers is equal.

Answer 13

Japt, 14 13 bytes

2õ@ò3 x_â ʶX

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Explanation

2õ                :Range [1,2]
  @               :Pass each X through a function
   ò3             :  Split input to arrays of length 3
       _          :  Pass each through a function
        â         :    Remove duplicates
          Ê       :    Get length
           ¶X     :    Test for equality with X
      x           :  Reduce by addition

Answer 14

Haskell, 90 bytes

g[]=[]
g(a:b:c:x)=(sum$map fromEnum[a==b,a==c,b==c]):g x
f x=[sum[1|y<-g x,y==n]|n<-[3,1]]

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Looks a bit awkward...

Answer 15

Python 2, 77 72 65 bytes

lambda a:map([len(set(t))for t in zip(*[iter(a)]*3)].count,(1,2))

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7 bytes saved via a clever trick from xnor

Answer 16

Retina, 23 bytes

S2,3,` 
%Cq`\S+
*\C`1
2

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Explanation

S2,3,` 

Split the input at every 3rd space starting at the (0-based) 2nd, i.e. split the input into groups of three.

%Cq`\S+

On each line (%) count the number (C) of unique (q) values (\S+).

*\C`1

Count the number of 1s and print them with a trailing linefeed (\), but do so in a dry-run (*) so that we don't lose the previous result.

2

Count the number of 2s (and print them automatically).

Answer 17

J, 16 15 bytes

-1 byte thanks to cole!

1#.1 2=/_3#@=\]

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Pretty much the same approach as the majority of solutions.

Explanation:

        _3    \]  - split the input into sublists of lenght 3
          #@~.    - for each triplet remove duplicates and take the length 
   1 2=/          - compare with 1 and 2
1#.               - add up

Answer 18

Wolfram Language (Mathematica), 49 bytes

Saved two bytes thanks to Martin Ender.

{#~Count~{_},#~Count~{_,_}}&@BlockMap[Union,#,3]&

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Answer 19

Elixir, 92 bytes

fn a->import Enum;c=map chunk(a,3),&(length uniq&1);{count(c,&(&1==1)),count(c,&(&1==2))}end

First, chunks the list into size length 3 chunk(a,3)

Secondly, it converts finds the length of each element, uniqified; map chunk(a,3),&(length uniq&1).

Finally, it returns an array consisting of the number of times the resulting list is equal to one count(c,&(&1==1)) and the number of times the resulting list is equal to two count(c,&(&1==2)).

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Answer 20

Ruby, 59 bytes

->a{[1,2].map{|x|a.each_slice(3).count{|y|x==y.uniq.size}}}

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Answer 21

Japt, 24 19 bytes

ò3 ®â l
[Uè¥1 Uè¥2]

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Answer 22

Stax, 14 bytes

ü┬─*HTÜ╫\Ä╢qm♥

Run and debug it

Answer 23

Prolog (SWI), 80 bytes

[A,B,C|T]-X/Y:-T-M/N,(A=B,B=C,X is M+1,Y=N;X=M,(A\=B,B\=C,Y=N;Y is N+1)).
_-0/0.

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Answer 24

Factor + pair-rocket, 58 bytes

[ 3 group 1 => 2 [ '[ cardinality _ = ] count ] with map ]

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Answer 25

Desmos, 84 bytes

l(r)=r.length
b=[l(k[3x-2...3x].unique)forx=[1...l(k)/3]]
f(k)=[l(b[b=1]),l(b[b=2])]

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Answer 26

Vyxal 14 13 bytes

3ẇƛ∪L;:1O$2Ox

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Explanation:

3ẇ             # 3-partition
  ƛ            # Open lambda map
   ∪L          # Length of sublist union
     ;         # Close lambda map
      :        # Duplicate for next operation
       1O      # Count 1
         $     # Swap
          2O   # Count 2
            x  # Print stack as list

Answer 27

C (GCC), 135 128 117 bytes

-7 bytes thanks to @ceilingcat

-11 more, again thanks to @ceilingcat

This is basically @ceilingcat's answer now, since he's improved it by more than 13 percent.

#define E(k)v[k-~i]==v[i
a;b;i;f(v,l)int*v;{for(a=b=i=0;i<l;i+=3)E()]&E(1)]?++a:E()]|E(1)]|E(1)+1]&&++b;*v=a;v[1]=b;}

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Unfortunately, I'm too lazy to write a new explanation for the most recent answer, but it's quite similar to the previous 128-byte one:

#define E(j,k)v[j+i]==v[k-~i]
a;b;i;f(v,l)int*v;{for(a=b=i=0;i<l;i+=3)E(,)&&E(,1)&&++a||(E(,)||E(,1)||E(1,1))&&++b;*v=a;v[1]=b;}

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f takes a vector (v) and its length (l) as arguments, and return its values in the first two elements of that vector. Explanation:

#define E(j,k)v[j+i]==v[k-~i] // k-~i == k-(-i-1) == k+i+1. an empty first argument equals +i, and an empty second argument equals -~i == i+1
a;b;i;
f(v,l)int*v;
{
    for(a=b=i=0;i<l;i+=3)
        E(,)&&E(,1)&&++a // if element i, i+1 and i+2 are all equal, increment a
        ||(E(,)||E(,1)||E(1,1))&&++b; // otherwise, if any two elements are equal, increment b
    *v=a; // set the first element in v to a
    v[1]=b; // set the second one to b
}

Answer 28

Nibbles, 9 bytes

.,2,`?.`/3@,`$$@

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Explanation

.    Map
,2    [1,2]
,     length of
`?     find indices in
.       map
`/ 3     split into chunks of 3
@         input
,        length of
`$        deduplicate
$          the chunk
@       of the number (1 or 2)

Answer 29

Elm, 108 106 bytes

import List.Extra as E
m=List.map
f x=m(\j->E.count((==)j)<|m(List.length<<E.unique)<|E.groupsOf 3 x)[1,2]

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Answer 30

Raku, 30 bytes

(+«*.rotor(3)».Set).Bag{1,2}

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• *.rotor(3) breaks up the input list into groups of three elements.
• ».Set converts each of those groups into a Set.
• +« converts each of those sets into its size.
• .Bag makes a Bag (a set with multiplicity) out of those sizes.
• {1, 2} looks up in that Bag the number of times sets with sizes of 1 and 2 were present.