19
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Given a list with number, output the ranges like this:

Input: [0, 5, 0] would become [0, 1, 2, 3, 4, 5, 4, 3, 2, 1, 0].

This is mapping a range through the array, so we first have to create the range [0, 5], which is [0, 1, 2, 3, 4, 5]. After that, we use the 5 to create the range [5, 0]. Appended at our previous range, this gives us:

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

Let's observe a test case with two same digits next to each other:

[3, 5, 5, 3], ranges:

[3, 5] = 3, 4, 5
[5, 5] = 5 (actually [5, 5] due to overlapping)
[5, 3] = 5, 4, 3

So this would give us [3, 4, 5, 5, 4, 3].

Some other test cases:

[1, 9] > [1, 2, 3, 4, 5, 6, 7, 8, 9]
[10, -10] > [10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5, -6, -7, -8, -9, -10]
[3, 0, 0, -3] > [3, 2, 1, 0, 0, -1, -2, -3]
[1, 3, 5, 7, 5, 3, 1, -1, -3] > [1, 2, 3, 4, 5, 6, 7, 6, 5, 4, 3, 2, 1, 0, -1, -2, -3]

Input will always have at least 2 integers.

Shortest answer wins!

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  • 3
    \$\begingroup\$ Related. Related. \$\endgroup\$ – Martin Ender Mar 5 '16 at 14:22
  • 1
    \$\begingroup\$ In what way are input and output related? What constitutes a valid input? \$\endgroup\$ – flawr Mar 5 '16 at 14:31

20 Answers 20

21
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05AB1E, 1 byte

Ÿ

Try it online!

How it works

It's a built-in.

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  • 18
    \$\begingroup\$ Do you have a dictionary of all built-ins in all esolangs in your head, or what? ;) \$\endgroup\$ – ETHproductions Mar 5 '16 at 17:51
  • 2
    \$\begingroup\$ Well, thank you for using osabie :P \$\endgroup\$ – Adnan Mar 5 '16 at 18:14
  • 7
    \$\begingroup\$ Why does it even have a built-in for this? \$\endgroup\$ – Neil Mar 5 '16 at 20:24
  • \$\begingroup\$ There should be a compilation of all 0byte and 1byte (maybe even 2byte) programs that do stuff. \$\endgroup\$ – CalculatorFeline Mar 6 '16 at 15:56
  • 2
    \$\begingroup\$ @Neil It's basically an inclusive range function, it's really not that spectacular. \$\endgroup\$ – Adnan Mar 6 '16 at 19:53
5
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Javascript, 99 95 93 bytes

4 6 bytes off thanks @Neil.

a=>a.reduce((x,y)=>x.concat(b.map?b=y:[...Array(y<b?b-y:y-b||1)].map(_=>b+=y<b?-1:y>b)),b=[])

f=
a=>a.reduce(
    (x,y)=>
        x.concat(
            b.map?b=y
            :[...Array(y<b?b-y:y-b||1)]
                .map(_=>b+=y<b?-1:y>b)
        )
    ,b=[])


G.addEventListener('click',_=>O.innerHTML=f(JSON.parse(I.value)));
<input id=I value="[3,5,5,3]"><button id=G>Go</button><pre id=O>

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  • 1
    \$\begingroup\$ Save 3 bytes by using y<b?b-y:y-b||1. Save another byte by using y>b||y-b&&-1. \$\endgroup\$ – Neil Mar 5 '16 at 20:15
  • \$\begingroup\$ @Neil. Good ones!! Thanks :) \$\endgroup\$ – removed Mar 5 '16 at 20:24
  • 1
    \$\begingroup\$ Actually y<b?-1:y>b is better still. \$\endgroup\$ – Neil Mar 5 '16 at 23:45
5
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JavaScript (SpiderMonkey 30+), 81 76 bytes

([n,...a])=>[n,...[for(i of a)for(j of Array(i<n?n-i:i-n||1))n+=i<n?-1:i>n]]

Tested in Firefox 44. Uses ES6's awesome argument destructuring capabilities and ES7's array comprehensions (which have sadly been removed from the ES7 spec).

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  • \$\begingroup\$ Doesn't work on [3, 0, 0, -3]. I fixed the RangeError and saved 10 bytes but it still doesn't work: ([n,...a],z=[n])=>z.concat([for(i of a)for(j of[...Array((r=n<i)?i-n-1:n-i-1),0])i=r?++n:--n]) \$\endgroup\$ – Neil Mar 5 '16 at 20:09
  • \$\begingroup\$ Sorry, I meant ([n,...a])=>[n].concat([for(i of a)for(j of[...Array((r=n<i)?i-n:n-i)])i=r?++n:--n]) of course. \$\endgroup\$ – Neil Mar 5 '16 at 20:26
  • \$\begingroup\$ @Neil Fixed, with a bunch more golfed off in the process \$\endgroup\$ – ETHproductions Mar 6 '16 at 1:27
4
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JavaScript (ES6) 66 72

A recursive function that repeatedly adds values inside the array to fill the gaps between near numbers

f=l=>l.some((x,i)=>(z=l[i-1]-x)*z>1&&l.splice(i,0,x+z/2|0))?f(l):l

Test

f=l=>l.some((x,i)=>(z=l[i-1]-x)*z>1&&l.splice(i,0,x+z/2|0))?f(l):l

console.log=x=>O.textContent+=x+'\n'

;[[1,9],[10,-10],[3,0,0,-3],[1, 3, 5, 7, 5, 3, 1, -1, -3]]
.forEach(t=>console.log(t+' -> ' +f(t)))
<pre id=O></pre>

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3
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C, 120 + 12 = 132 bytes

i,j,k;f(a,n)int*a;{a[0]--;for(i=0;i<n-1;i++)for(k=0,j=a[i]-a[i+1]?a[i]:a[i]-1;j-a[i+1];)printf("%i ",j+=a[i+1]>j?1:-1);}

Example call:

f(a,sizeof(a)/4);        // I've added 12 bytes because of ",sizeof(a)/4"

Test live on ideone.

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3
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Python 2, 77 bytes

lambda n:n[0:1]+sum([range(x,y,2*(y>x)-1)[1:]+[y]for(x,y)in zip(n,n[1:])],[])

Try it online

Thanks to Neil, DenkerAffe, and Erwan for pointing out improvements that I missed

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  • \$\begingroup\$ Surely the +1 is unnecessary? \$\endgroup\$ – Neil Mar 5 '16 at 20:21
  • \$\begingroup\$ why not go with lambda n:n[0:1]+sum([range(x,y,[1,-1][y+1<x])[1:]+[y]for(x,y)in zip(n,n[1:])],[])? saves some bytes. \$\endgroup\$ – Denker Mar 5 '16 at 22:46
  • \$\begingroup\$ I was very tired while writing this :) Answer first, improve later. \$\endgroup\$ – Mego Mar 5 '16 at 23:43
  • \$\begingroup\$ i think you can replace [1,-1][y+1<x] by 2*(y>x)-1 (also i don't understand why you use y<=x and not simply y<x ) \$\endgroup\$ – Erwan Mar 9 '16 at 9:04
  • \$\begingroup\$ n[0:1] is equivalent to n[:1]. \$\endgroup\$ – Jonathan Frech Oct 7 '17 at 16:18
3
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Perl, 47 bytes

Includes +3 for -p (code contains $' so space and - count too)

Give the list of numbers on STDIN:

fluctuating.pl <<< "3 5 5 3"

fluctuating.pl:

#!/usr/bin/perl -p
($n=$&+($'<=>$&))-$'&&s/\G/$n / while/\S+ /g

The temporary variable and all these parenthesis feel suboptimal...

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  • \$\begingroup\$ It looks like you posted the wrong answer : it doesn't seem to be working and that $' you mentioned isn't in the code... \$\endgroup\$ – Dada Sep 7 '16 at 20:33
  • \$\begingroup\$ @Dada: Yes, again pasted an old untested version of the code instead of the fixed one. Thanks and fixed \$\endgroup\$ – Ton Hospel Sep 7 '16 at 22:47
2
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Haskell, 63 55 bytes

g(a:b:r)=[a|a==b]++[a..b-1]++[a,a-1..b+1]++g(b:r)
g x=x

Usage example: g [3,5,5,3] -> [3,4,5,5,4,3].

It's a modification of my answer to a related challenge. Again, the main work is done by concatenating the list from a upwards to b-1 and from a downwards to b+1 (where one list will be empty) and a recursive call. To handle the a==b case where both lists are empty, we prepend [a|a==b] which evaluates to [a] if a==b and [] otherwise.

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2
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R, 86 82 75 bytes

function(x)rep((y<-rle(unlist(Map(seq,head(x,-1),x[-1]))))$v,pmax(1,y$l-1))

saved 4 bytes using rep not rep.int (code golf not performance!) saved another 7 bytes by using built-in partial matching when using $ (and collapsing function definition to 1 line

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  • \$\begingroup\$ I think (y=...) rather than (y<-...) is also valid, and one byte less. \$\endgroup\$ – Giuseppe Oct 6 '17 at 14:51
2
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Ruby, 116 82 bytes

->n{o,*m=n;o=[o];m.zip(n).map{|t,u|o+=[[u],[*u+1..t],[*t..u-1].reverse][t<=>u]};o}

My first ever golf.

Edit: Thanks manatwork for the awesome suggestions.

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  • \$\begingroup\$ No need to assign to variable, anonymous proc is enough; no need to put parenthesis around formal parameter; taking out the array's first element is shorter with parallel assignment and splat; map's code block can take the array as multiple parameters: ->n{o,*m=n;o=[o];m.zip(n).map{|t,u|o+=u==t ?[u]:(u<t ?[*u+1..t]:[*t..u-1].reverse)};o}. Otherwise nice first golf. \$\endgroup\$ – manatwork Mar 8 '16 at 7:17
  • \$\begingroup\$ Picking from a 3 element array by spaceship operator is shorter than 2 ternary operators: [[u],[*u+1..t],[*t..u-1].reverse][t<=>u]. \$\endgroup\$ – manatwork Mar 8 '16 at 7:40
2
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Japt, 12 bytes

Saved 16 bytes thanks to @ETHproductions!

ä!õ ËsE©DÊ>1

Test it online

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1
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Perl 6, 94 bytes

I'm not super happy with this right now, I'll probably take another shot later

{reduce {|@^a[0..*-2],|@^b},map {@_[0]!= @_[1]??(@_[0]...@_[1])!!(@_[0],@_[1])},.rotor(2=>-1)}
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1
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PHP 5.4, 86 bytes

This is meant to be used as an included file, that returns the result.

The values are passed as commandline parameters.

<?for($i=1;$i<$argc-1;$R=array_merge($R?:[],range($argv[$i++],$argv[$i++])));return$R;

Not exactly pretty or anything, but does the job.

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1
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Python 3, 76 bytes

First attempt at a Python answer. The basic idea is to repeatedly identify pairs in the sequence where the difference is larger than a step and insert one (and only one) additional element to complete the sequence in the right direction. Repeat until all differences between consecutive elements are between +1 and -1.

d=diff
while any(d(x)**2>1):i=argmax(d(x)**2);x[:i+1]+=[x[i]+sign(d(x)[i])]

Try it online!

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0
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Lua, 156 Bytes

A function that takes an array in parameter and return the extended array.

function f(t)r={}for i=2,#t
do x,y=t[i-1],t[i]r[#r+1]=x==y and x or nil
z=x>y and-1or 1
x=x==r[#r]and x+z or x
for j=x,y,z
do r[#r+1]=j end end
return r end

Ungolfed and explanations

function f(t)
  r={}                        -- Initialise an empty array
  for i=2,#t                  -- Iterate over the parameter array
  do
    x,y=t[i-1],t[i]           -- x and y are shorter names than t[i-1]
    r[#r+1]=                  -- when there's a range like [5,5]
      x==y and x or nil       -- put this number once in the array
    z=x>y and-1or 1         -- determine the step value
    x= x==r[#r]               -- prevent repeating the last value of r
          and x+z or x        -- by incrementing/decrementing x
    for j=x,y,z               -- iterate from t[i-1] to t[i] by step z (-1 or 1)
    do
      r[#r+1]=j               -- put j into the array r
    end
  end
  return r                    -- return the extended array
end

For ease of use, you can use the following function to print the array returned by f().

function printArray(t)
  print("["..table.concat(t,",").."]")
end

When testing this submission, you can call it like:

printArray(f( {0,5,0,3,4,4,7,3,-3} ))
> [0,1,2,3,4,5,4,3,2,1,0,1,2,3,4,4,5,6,7,6,5,4,3,2,1,0,-1,-2,-3]
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0
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Mathcad, 62 "bytes"

enter image description here

As Mathcad uses a 2D "whiteboard" and special operators (eg, summation operator, integral operator), and saves in an XML format, an actual worksheet may contain several hundred (or more) characters. For the purposes of Code Golf, I've taken a Mathcad "byte count" to be the number of characters or operators that the user must enter to create the worksheet.

Converting the function definition to a straight program, and replacing the variable lst with a single character name, gives a total of 62 "bytes". With the function, using a single character rather than the full name, this increases to 65 "bytes" for the definition and a further 4 "bytes" for each call (assuming that creation of the list itself isn't included in the overall byte count (Using Mathcad's built-in tables is another way of inputting the list).

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0
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PHP, 144 bytes

function f($r){$m=[];for($i=0;++$i<count($r);){$d=array_pop($m);$n=$r[$i];$p=$r[$i-1];$m=array_merge($m,$p==$n?[$p,$n]:range($p,$n));}return$m;}
Exploded view
function f($r) {
  $m = [];
  for ($i=0; ++$i < count($r); ) {
    $d = array_pop($m);
    $n = $r[$i];
    $p = $r[$i-1];
    $m = array_merge($m, $p==$n ? [$p,$n]
                                : range($p,$n));
  }
  return $m;
}
Input / function call
f([ bound1, bound2, bound3, ... ]);
Output
[int, int, int, int, ...]

It's messy and chunky, and I'll try to optimize it later. It creates a range() from each pair of adjacent value pairs, then stitches them together (after poping off the end of the previous cumulative Array).

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0
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Perl6, 21

.join is short for $_.join

say EVAL .join: "..."

Test (rakudo)

perl6 -MMONKEY-SEE-NO-EVAL -e'say EVAL @*ARGS.join: "..."' 1 3 5 7 5 3 1 -1 -3

Output

(1 2 3 4 5 6 7 6 5 4 3 2 1 0 -1 -2 -3)
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0
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Jelly, 10 bytes

,r⁼?2\Ṗ;¥/

Try it online!

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-1
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R, 74 bytes

Another R solution

function(x){b=x[1];z=c();for(a in x[-1]){z=c(z,c(b:(a-sign(a-b))));b=a};z}

Try it online!

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  • \$\begingroup\$ this doesn't quite work as the last value seems to be missing... \$\endgroup\$ – Giuseppe May 17 '18 at 20:19

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