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

APL, 4 characters

/⍨⍳⎕

How it works:

⎕ reads user input. As for output, APL by default prints the result from every line.

⍳n is the integers from 1 to n. Example: ⍳3←→ 1 2 3

/ means replicate. Each element from the right argument is repeated as many times as specified by its corresponding element from the left argument. Example: 2 0 3/'ABC'←→ 'AACCC'

⍨ is the commute operator. When it occurs to the right of a function, it modifies its behaviour, so it either swaps the arguments (A f⍨ B ←→ B f A, hence "commute") or provides the same argument on both sides (f⍨ A ←→ A f A, a "selfie"). The latter form is used in this solution.

---

Bonus:

6-∊⌽⍳¨⍳⎕ (8 characters, thanks @phil-h)

⍳5 (iota five) is 1 2 3 4 5.

⍳¨ ⍳5 (iota each iota five) is (,1)(1 2)(1 2 3)(1 2 3 4)(1 2 3 4 5), a vector of vectors. Each (¨) is an operator, it takes a function on the left and applies it to each item from the array on the right.

⌽ reverses the array, so we get (1 2 3 4 5)(1 2 3 4)(1 2 3)(1 2)(,1).

∊ is enlist (a.k.a. flatten). Recursively traverses the argument and returns the simple scalars from it as a vector.

Answer 2

Ruby (recursive), 41 bytes * 0.5 = 20.5

def n(n,i=1);i>n ?[]:n(n,i+1)+[*i..n];end

Or using a lambda (as recommended by histocrat and Ventero): 34 bytes * 0.5 = 17

r=->n,i=n{i>0?[*i..n]+r[n,i-1]:[]}

(call using r[argument])

Answer 3

Pyth, 9 bytes * 0.5 = 4.5

smrhQhdUQ

With help from @FryAmTheEggman

Try it online.

---

Explanation

s             reduce + on list
 m            map
  rhQhd       lambda d: reversed(range(d+1, Q+1)), over
       UQ     range(Q)

where Q is the input.

Answer 4

Haskell, 31 characters = 15.5 score

f n=[y|x<-[n,n-1..1],y<-[x..n]]

27 characters without the bonus

f n=[x|x<-[1..n],_<-[1..x]]

Beaten by Proud Haskeller

Answer 5

C, 22 = 44 bytes * 0.5

The function h takes two parameters. The first is an int specifying n. The second is an int* which is the output buffer.

h(n,o)int*o;{for(n&&h(~-n,o+=n);*--o=n--;);}

Test program

main(){
int wow[999],*i;
memset(wow,0,sizeof(wow));
h(6, wow);
for(i=wow;*i;i++)printf("%d ", *i);
}

Answer 6

Pyth - 15 10 * .5 = 5

smr-QdhQUQ

Try it online.

Expects input on stdin. Independently discovered algorithm. Thanks @Sp3000 for helping me stick the last Q in there :P Also, irony? XD

Explanation:

Q=eval(input())       : implicit
s                     : The sum of...
 m      UQ            : map(...,range(Q))
  r-QdhQ              : range(Q-d,Q+1)

Answer 7

Python 2, 53 bytes * 0.5 = 26.5

i=n=input()
x=[]
while i:x+=range(i,n+1);i-=1
print x

Shamelessly borrowed @VisualMelon's idea

Answer 8

Haskell, 34 bytes * 0.5 = 17

0%n=[]
i%n=[i..n]++(i-1)%n
g n=n%n

That's the first time I've ever used Haskell for golfing. Call with g <number>.

Answer 9

CJam, 12 15 bytes * 0.5 = 7.5

li_,f{),f-W%~}`

This is full STDIN-to-STDOUT program. It concatenates increasing suffixes of the 1 ... n range, which ensures that no two adjacent numbers are identical.

Test it here.

Answer 10

C# - 81 (161bytes * 0.5)

Simple job in C#, hopefully gets the no-neibouring-numbers bonus. Reads an int from stdin, writes out an array like the example to stdout.

class P{static void Main(){int n=int.Parse(System.Console.ReadLine()),m=n-1,i;var R="["+n;for(;m-->0;)for(i=m;i++<n;)R+=", "+i;System.Console.WriteLine(R+"]");}}

More readable:

class P
{
    static void Main()
    {
        int n=int.Parse(System.Console.ReadLine()),m=n-1,i;
        var R="["+n;
        for(;m-->0;)
            for(i=m;i++<n;)
                R+=", "+i;
        System.Console.WriteLine(R+"]");
    }
}

Examples output:

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

Answer 11

Bash + coreutils, 28/2 = 14

Shamelessly stealing @pgy's idea and golfing:

seq -f"seq %g $1" $1 -1 1|sh

---

Pure bash (no coreutils), 30/2 = 15

Eval, escape and expansion hell:

eval eval echo \\{{$1..1}..$1}

Answer 12

GolfScript (14 bytes * 0.5 = score 7)

 ~:x,{~x),>~}%`

Online demo

I think this is probably similar to some existing answers in that it builds up the array concat( [n], [n-1, n], [n-2, n-1, n], ..., [1, 2, ..., n] )

Sadly I wasn't able to golf any further the arguably more elegant:

~:x]{{,{x\-}/}%}2*`

which puts the input x into an array and then twice applies {,{x\-}/}%, which maps each element in an array to a count down of that many elements from x.

Answer 13

JavaScript, ES6, 41 bytes

f=i=>[...Array(i).fill(i),...i?f(--i):[]]

This creates a function f which can be called like f(6) and it returns the required array.

This uses a recursive approach, where each iteration creates an array of i elements all valued i and concatenates an array returned by f(i-1) with stopping condition of i==0.

Works on latest Firefox.

Answer 14

vba, 76*0.5=38

Sub i(q)
For Z=1 To q:For x=q To Z Step -1:Debug.?x;",";:Next:Next
End Sub

Answer 15

Haskell, 14 = 28 bytes / 2

f n=n:[1..n-1]>>= \r->[r..n]

example output:

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

24 bytes without the bonus:

f n=[1..n]>>= \r->[r..n]

Answer 16

Haxe, 53 bytes

function l(v)return[for(i in 0...v)for(j in 0...i)i];

Works with l(6); because of array comprehension.
Test online http://try.haxe.org/#741f9

Answer 17

perl ,26 bytes

for(1..$n){print"$_ "x$_;}

Answer 18

R, 44 *.5 = 22

f=function(n){r=0;for(i in 1:n)r=c(r,n:i);r}

A quick test

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

Answer 19

JavaScript, ES6, 66 bytes * 0.5 = 33

f=i=>(g=n=>[...Array(n).fill().map((v,x)=>i-x),...n?g(n-1):[]])(i)

Building on Optimizer's recursive approach, we can build descending runs of decreasing length, like [4,3,2,1, 4,3,2, 4,3, 4].

Instead of making same-value subarrays with Array(i).fill(i), we make undefined-filled subarrays of the appropriate length with Array(n).fill() and then change the values to a descending run using .map((v,x)=>i-x). Also, we define and recurse over an inner function g; the outer function f exists only to store the value of i while g recurses.

Answer 20

T-SQL, 176 * 0.5 = 88

Since you seemed to miss the T-SQL @Optimizer, here it is in all it's verbose glory :).

A couple of function options, a Scalar and a Inline Table Valued function. The Scalar function uses while loops to recurse and returns a string of numbers, where the Inline Table Valued function uses a recursive CTE for a sequence and returns a table. Of course these will never be competitive, so I haven't spent a lot of time golfing.

Inline Table Valued Function, 176 * .5

CREATE FUNCTION F(@ INT)RETURNS TABLE RETURN WITH R AS(SELECT @ N UNION ALL SELECT N-1FROM R WHERE N>0)SELECT B.N FROM R CROSS APPLY(SELECT TOP(R.N)N FROM R A ORDER BY N DESC)B

Called as follows

SELECT * FROM dbo.F(5)

SQLFiddle example

Scalar Function, 220 * .5

CREATE FUNCTION G(@ INT)RETURNS VARCHAR(MAX)AS BEGIN DECLARE @S VARCHAR(MAX),@N INT=1,@I INT,@C INT WHILE @N<=@ BEGIN SELECT @I=@N,@C=@ WHILE @C>=@I BEGIN SELECT @S=CONCAT(@S+',',@C),@C-=1 END SET @N+=1 END RETURN @S END

Called as follows

SELECT dbo.G(5)

SQLFiddle example

Answer 21

Mathematica, 34 * 0.5 = 17

f=Join@@Table[x,{y,#},{x,#,y,-1}]&

Answer 22

JavaScript(readable), 131 bytes

I'm new to Code Golf so this isn't the best

function f(n) {
    var arr = [];
    for(var i = 1; i <= n; i++) {
        for(var j = 0; j < i; j++) {
            arr.push(i);
        }
    }
    return arr;
}

JavaScript(less readable), 87 bytes

Minified using jscompress.com

function f(e){var t=[];for(var n=1;n<=e;n++){for(var r=0;r<n;r++){t.push(n)}}return t}

Answer 23

TECO, 25 bytes * 0.5 = 12.5

a\+1%a%b<qauc-1%b<-1%c=>>

The above barely beats the non-bonus version at 13 bytes:

a\%a<%b<qb=>>

Answer 24

C#, 114 99 * 0.5 = 49.5 bytes

(With a little help from VisualMelon's answer) Edit: and James Webster's comment

int[]A(int n){int m=n,i,c=0;var a=new int[n*(n+1)/2];while(m-->0)for(i=m;i++<n;)a[c++]=i;return a;}

Ungolfed:

int[] FooBar(int n)
{
    int altCounter = n, i, arrayCounter = 0;
    var returnArray = new int[n * (n + 1) / 2];
    while(m-->0)
        for(i = altCounter; i++ < n; )
            returnArray[arrayCounter++]=i;
    return returnArray;
}

---

There is an unsafe version that I shamelessly took from feersum's C answer, but I'm not 100% sure it fits within the rules since you have to allocate the memory before calling the method.

C# (unsafe), 82 * 0.5 = 41 bytes

unsafe void A(int n,int*p){int*z=p;int m=n,i;while(m-->0)for(i=m;i++<n;)z++[0]=i;}

Called as follows:

int n = 5, length = (int)((n / 2f) * (n + 1));
int* stuff = stackalloc int[length];
int[] stuffArray = new int[length];
A(n, stuff);
System.Runtime.InteropServices.Marshal.Copy(new IntPtr(stuffArray), stuffArray, 0, stuffArray.Length);
//stuffArray == { 5, 4, 5, 3, 4, 5, 2, 3, 4, 5, 1, 2, 3, 4, 5 }

Per VisualMelon's suggestion (thanks!), the unsafe code can be re-made with safe code which reduces the size even further! Still poses the question if the creation of the final result array is allowed to be done outside of the method.

C#, 72 * 0.5 = 36 bytes

void A(int n,int[]p){int z=0,m=n,i;while(m-->0)for(i=m;i++<n;)p[z++]=i;}

Answer 25

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 26

C#, 116 115 + 33 = 148 bytes

Not the shortest code, but... it works anyway :P

int[]l(int m){List<int>i=new List<int>();for(int j=1;j<=m;j++){for(int x=0;x<j;x++){i.Add(j);}}return i.ToArray();}

Requires this at the top of the file (33 bytes):

using System.Collections.Generic;

Un-golfed version:

int[] RepatedNumberList(int m)
{
    List<int> intList = new List<int>();
    for (int j = 1; j <= m; j++)
    {
        for (int x = 0; x < j; x++)
        {
            intList.Add(j);
        }
    }
    return initList.ToArray();
}

Answer 27

Bash with seq, expr and xargs = 59 / 2 = 29.5

Save it and run with the number as the first argument.

seq $1|xargs -n1 seq|xargs -n1 expr $1 + 1 -|sed 1d;echo $1

Answer 28

J, 23 * 0.5 = 11.5

   f=.-;@(<@|.@i."0@>:@i.)
   f 5
5 4 5 3 4 5 2 3 4 5 1 2 3 4 5

J, 11

   f=.#~@i.@>:
   f 5
1 2 2 3 3 3 4 4 4 4 5 5 5 5 5

Answer 29

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 30

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 31

Tcl, 36 bytes

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

Try it online!

Answer 32

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 33

Jelly, 3 × 0.5 = 1.5 bytes

rRF

Try it online!

Explanation

rRF
r      Range from {input} to 
 R       each number from 1 to {input}
  F    Flatten

For example, with input 3, the rR gives us a range from 3 to 1 (i.e. [3, 2, 1], then a range from 3 to 2 (i.e. [3, 2]), then a range from 3 to 3 (i.e. [3, 3]). Flattening that gives [3, 2, 1, 3, 2, 3]), which meets the problem specifications and has no adjacent numbers.

Answer 34

K (oK), 4 bytes

Solution:

&!1+

Try it online!

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 35

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.

Try It Online

Answer 36

Pyth, 6 bytes

smrQdU

Try it online!

Answer 37

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
}

Try it online!

Pretty straightforward solution and probably pretty bad.

Answer 38

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 39

GolfScript, 14 bytes

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

Try it online!

Answer 40

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 41

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 42

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 43

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 44

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 45

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 46

CoffeeScript, 57

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

Answer 47

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.