Create an array with repeated numbers

Question

Challenge

Your task in this question is to write a program or a named function which takes a positive integer n (greater than 0) as input via STDIN, ARGV or function arguments and outputs an array via STDOUT or function returned value.

Sounds simple enough ? Now here are the rules

• The array will only contain integers from 1 to n
• Each integer from 1 to n should be repeated x times where x is the value of each integer.

For example:

Input:

5

Output:

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

The array may or may not be sorted.

This is code-golf so winner is shortest code in bytes.

Bonus

Multiply your score by 0.5 if no two adjacent integers in your output array are same.

For example for n = 5, one such configuration would be

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

Answer 1

JavaScript (ES6) 29 (58 * 0.5)

Edit remove ; thx @Optimizer

Q=o=>(m=>{for(n=o,r=[];n>m||++m<(n=o);)r.push(n--)})(0)||r

Test in FireFox/FireBug console

Q(9)

Output

[9, 8, 7, 6, 5, 4, 3, 2, 1, 9, 8, 7, 6, 5, 4, 3, 2, 9, 8, 7, 6, 5, 4, 3, 9, 8, 7, 6, 5, 4, 9, 8, 7, 6, 5, 9, 8, 7, 6, 9, 8, 7, 9, 8, 9]

Ungolfed

Q=o=>{
  for(m=0,r=[];m<o;++m)
    for(n=o;n>m;)
      r.push(n--);
  return r
}

Answer 2

ECMAScript6, 67 * 0.5 = 33.5 bytes

f=n=>{a=[],b=0;while(c=n+b,n--){while(c-b)a.push(c--);b++}return a}

Pretty happy with this one...It's about a quarter the size of my original.

f(4) returns:

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

Old answer:

f=i=>{a=b=Array;while(i)a=a.concat(b.apply(null,b(i)).map(e=>i)),i--;return a}

This is my first shot at code golf...I still want to get that 0.5x bonus. Any suggestions are welcomed!

Called with f(n).

Answer 3

C#, 108 bytes * 0.5 = 54

List<int> f(int n){var r=new List<int>();int m=n-1,i;r.Add(n);for(;m-->0;)for(i=m;i++<n;)r.Add(i);return r;}

Thanks to VisualMelon for doing the hard work! I thought I'd try to squeeze it down as much as possible.

(114 bytes * 0.5 = 57, if you insist on using .ToArray() to return int[])

Answer 4

K (oK), 4 bytes

Solution:

&!1+

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Examples:

&!1+5
1 2 2 3 3 3 4 4 4 4 5 5 5 5 5
&!1+6
1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6

Explanation:

&!1+ / the solution
  1+ / add one to the input, 1+5 => 6
 !   / til, !6 => 0 1 2 3 4 5
&    / where, &0 1 2 3 4 5 => 1 2 2 3 3 3...

Answer 5

Python, 53 bytes ÷ 2 = 26.5

lambda n:[n-i%-~n%(i//n-~n)for i in range(n*~n//2,0)]

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Python 2 version, 51 bytes ÷ 2 = 25.5

lambda n:[n-i%-~n%(i/n-~n)for i in range(n*~n/2,0)]

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This uses ranges like many other answers here but groups them in pairs (longest,shortest);(2nd longest,2nd shortest) etc. That way we can use a closed form expression inside a flat loop.

Answer 6

Factor + grouping.extras, 35 * 0.5 = 17.5 bytes

[ [1,b] tail-clump reverse concat ]

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           ! 5
[1,b]      ! { 1 2 3 4 5 }
tail-clump ! { { 1 2 3 4 5 } { 2 3 4 5 } ... }
reverse    ! { { 5 } { 4 5 } { 3 4 5 } ... }
concat     ! { 5 4 5 3 4 5 2 3 4 5 1 2 3 4 5 }

Answer 7

JavaScript (V8), 44 bytes

y=x=>x-1?y(x-1).concat(Array(x).fill(x)):[1]

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Could probably be golfed more and probably similar to some other answers but might as well post it.

Answer 8

Vyxal, 8 6 4 bytes

ƛẋ;f

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Explained

ƛẋ;f
ƛ    # mapping lambda
 ẋ   # Repeat
  ;  # end lambda
   f # flatten

Answer 9

Raku, 25 bytes * 0.5 = 12.5

{$_,|flat (1..$_-1)X..$_}

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This generates the list consisting of the input number n, and the flattened list of ranges 1 .. (n-1), 2 .. (n-1), ..., n-2 .. n-1, n-1 .. n-1.

Answer 10

Japt, 5 bytes * 0.5 = 2.5

õõU c

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õõU c #
õ     # For all numbers X in the range [1, input]:
 õU   #  Generate the range [input, X]
    c # Flatten

A cheatier 3 byte version produces the same visible output, but that's only because Japt's default string representation of 2D arrays isn't very clear. Using the -Q flag reveals the real output.

Answer 11

K (ngn/k), 12 bytes

,/{x#x}'1+!:

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Short and sweet.

Explanations:

,/{x#x}'1+!:  Main function. Takes implicit input x via Right (:)
          !   Create a range between 0 to x (exclusive)
        1+    + 1 to each of them
       '      For each of the numbers in the range
  {   }       Execute a function that
     x        Takes the number as implicit argument x
   x#         And duplicate by themselves' amount of times
,/            Join together

Answer 12

Uiua, 9 bytes, 5 characters

▽.+1⇡

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Code explained:

⇡   # Create an array of integers from 0 to [input]
+1  # Add 1 to each element
.   # Duplicate the array
▽   # Create [element value] amount of copies of each element of the array

Answer 13

Python 3.8 (pre-release), 52 × 0.5 = 26 bytes

lambda n:[j+n for i in range(n)for j in range(-i,1)]

A simple double list comprehension solution.

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Answer 14

Perl, 57 * 0.5 = 28.5

This entry is a subroutine named "l" (for "list"?)

sub l{$d=$n=shift;while($d){for($d--..$n){@a=(@a,$_)}}@a}

You can test it like this:

$"=", ";
@a=l(3);
print "[@a]\n"; # prints "[3, 2, 3, 1, 2, 3]"
@a=();          # needed because the function does not re-initialize @a
@a=l(5);
print "[@a]\n"; # prints "[5, 4, 5, 3, 4, 5, 2, 3, 4, 5, 1, 2, 3, 4, 5]"

A prior solution is shorter (45 characters), but doesn't print the output prettily so it was disqualified

$d=$n=<>;while($d){for($d--..$n){print"$_ "}}

For an input of 3, this one prints (with a trailing space):

3 2 3 1 2 3 

Answer 15

Perl 48 * 0.5 = 24

$i=<>;$"=", ";@a=map{-$_..$i}-$i..-1;print"[@a]"

Test run:

$ ./repeated.pl <<< 5
[5, 4, 5, 3, 4, 5, 2, 3, 4, 5, 1, 2, 3, 4, 5]

Answer 16

Erlang, 56 Bytes * 0.5 = 28

f(I)->L=lists,L:flatten([L:seq(X,I)||X<-L:seq(I,1,-1)]).

Using a list comprehension, reverses the list using lists:seq(UpperBound, 1, -1) and for each element, creates a sequence using lists:seq(CurrentValue, UpperBound). The eventual list is then flattened.

Expanded:

f(I) ->
    lists:flatten([lists:seq(X, I) || X <- lists:seq(I, 1, -1)]).

Input: module_name:f(9).

Output: [9,8,9,7,8,9,6,7,8,9,5,6,7,8,9,4,5,6,7,8,9,3,4,5,6,7,8,9,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9]

Answer 17

Perl 5: 30 bytes / 2 = 15

Added one byte for the -p flag.

#!perl -p
$_=join$",map-$_..$',-$_..-//

Takes parameter on the input:

$ perl int.pl <<<5
5 4 5 3 4 5 2 3 4 5 1 2 3 4 5

Answer 18

Java (adjacent equals ok): 276 bytes

public class a{public static int[]b(final int c){final int d=(c+1)*c/2;final int[]e=new int[d];int f=1;int g=0;for(int i=0;i<d;i++){if(g==f){f++;g=0;;}
e[i]=f;g++;}
return e;}
public static void main(String[]args){final int q=new java.util.Scanner(System.in).nextInt();b(q);}}

Java (adjacent equals not ok): 442 * 0.5 = 221 bytes

public class a{private static int h=1;public static int[]j(final int k){final int[]l=new int[k+1];for(int i=1;i<=k;i++){l[i]=i;}
final int m=(k+1)*k/2;final int[]n=new int[m];for(int i=1;i<n.length;i++){n[i]=o(l,k);}
n[0]=k;return n;}
private static int o(final int[]p,final int q){if(h>q){h=1;}
while(p[h]==0){h++;}
p[h]=p[h]-1;return h++;}
public static void main(String[]args){final int q=new java.util.Scanner(System.in).nextInt();j(q);}}

Answer 19

CoffeeScript, 57

f=(n)->i=0;a=[];(j=0;a.push i while(j++<i))while(i++<n);a

Answer 20

JAGL Alpha 1.1 - 15 bytes

1Ti[r{d()+S*}/E

Explanation:

1Ti               Push 1, take input, and convert to integer
   [r             Increment and make a range
     {d()+S*}     Push a block that makes an array with x occurrences of x
             /E   Map over list and flatten

Answer 21

Matlab, 44 * 0.5 = 22

function r=f(n);r=[];for i=1:n;r=[i:n,r];end

f(4) == [4 3 4 2 3 4 1 2 3 4]

Javascript, 63 * 0.5 = 31.5

F=n=>{r=[m=n];while(--n){r=r.slice(0,m-n).concat(n,r)}return r}

F(4) == [ 4, 3, 2, 1, 4, 3, 2, 4, 3, 4 ]

Nothing fancy, just concatenating arrays.

Answer 22

Tcl, 36 bytes

time {time {puts $i} [incr i]} $argv

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Answer 23

Stax, 4 bytes * 0.5 = 2

R|]r

Run and debug it online! Note: To verify that is does indeed work, you'll need to step through the execution, as the values are printed as characters rather than numbers.

Explanation:

R|]r
R       Range [1..n].     Example: [1, 2, 3, 4]
 |]     List of suffixes: Example: [[1, 2, 3, 4], [2, 3, 4], [3, 4], [4]]
   r    Reverse.          Example: [[4], [3, 4], [2, 3, 4], [1, 2, 3, 4]]
        Implicit flatten and print (treating numbers as characters and concatenating).

Without the bonus (3 bytes):

R|]

or

R|[     (This does what you expect)

Answer 24

Java 8, 109 bytes (score: 54.5 bytes)

import java.util.*;

n->{List l=new Stack();l.add(n);for(int i=0,j;i<n-1;i++)for(j=i;j++<n;)l.add(j);return l;}

That's a lambda from int to List<Object>. The members are, of course, Integers.

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Answer 25

Golang, 80 bytes

func n(a int)(r[]int){for i:=0;i<=a;i++{for j:=0;j<i;j++{r=append(r,i)}};return}

Ungolfed and readable:

func n(a int) (r []int){
    for i := 0; i <= a; i++ {
        for j := 0; j < i; j++ {
            r = append(r, i)
        }
    }
    return
}

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Pretty straightforward solution and probably pretty bad.

Answer 26

Pyth, 17 bytes x 0.5 = 8.5

FbQ=+Y-SQSb)+eYPY

Possibly golfable. Explanation:

FbQ=+Y-SQSb)+eYPY
FbQ                For b in range(int(input()))
     Y             A variable containing an empty list
   =+              +=
      -            The set difference of
       SQ          [1, 2, ..., input()]
         Sb        [1, 2, ..., b]
           )       End function
            +eY    Append to the last element of Y
               PY  All but the last element of Y

So for example, putting 3 through the loop would yield [1,2,3,2,3,3], then the last five characters will transpose that final 3 to the beginning so no two numbers touch each other.

Answer 27

GolfScript, 14 bytes

~),1>{.[()]*}%

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Answer 28

Add++, 9 bytes

L,Rdz£XbF

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In most other languages, it might be worth going for the bonus by using things like prefixes etc. Not in add++.

Explained

L,Rdz£XbF
L,        # create a lambda that:
  R       # pushes the range [1...n] to the stack
   dz     # ... zips that with itself
     £X   # ... reduces each item by item repetition
       bF # ... and flattens that

Answer 29

brainf***, 98 bytes

[[->+>+<<]>>[-<<+>>]<-[[->+>+<<]>>[-<<+>>]<-<[-]<[->+>>+<<<]>>>[-<<<+>>>]<]<[->+>+<<]>>[-<<+>>]<-]

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Takes input as first cell on tape and outputs onto the tape. I'm not sure if this is allowed but input and output is really painful in brainf*** so I'm presuming it's allowed. The example on TIO has header and footer code to print the values out because I couldn't find a way to dump the memory into the output.

If you consider that each brainf*** character only contains 3 bits of information (as there are only 8 possible characters), then it's only 98 * (3/8) = 36.75 bytes of information.

I would put an explanation but I wrote the code yesterday and have forgotten how it works since then.

Answer 30

MathGolf, score: 2 (4 bytes \$\times \frac{1}{2}\$ bonus)

╒╒x─

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Explanation:

╒     # Push a list in the range [1, (implicit) input]
 ╒    # Map each value to an inner list in the range [1, value]
  x   # Reverse this list of lists
   ─  # Flatten
      # (after which the entire stack is output implicitly as result)