Collapse consecutive integers

Question

Related: Tell me how many math problems I have to do!

Challenge

Given a strictly positive strictly ascending integer list L and an integer 3 ≤ N ≤ length of L, replace the middle integers of L's consecutive integer runs of length ≥ N with a single dash -.

Rules

• Horizontal whitespace is irrelevant.
• You may optionally preserve the introducer, separator, and terminator characters of your language's default list format. See Format examples, below.

Data examples

All these examples use L = 3 5 6 7 8 10 11 12 14 16 17 18 19 20 21 22 24.

N = 3 → 3 5 - 8 10 - 12 14 16 - 22 24

N = 4 → 3 5 - 8 10 11 12 14 16 - 22 24

N = 5 → 3 5 6 7 8 10 11 12 14 16 - 22 24

N = 8 → 3 5 6 7 8 10 11 12 14 16 17 18 19 20 21 22 24

Format examples

For the inputs
 L = [3,5,6,7,8,10,11,12,14,16,17,18,19,20,21,22,24] and N = 3
all the below lines are examples of valid responses, both as actual lists and as strings:

[3,5,"-",8,10,"-",12,14,16,"-",22,24]
[3,5,-,8,10,-,12,14,16,-,22,24]
[3,5-8,10-12,14,16-22,24]
3,5-8,10-12,14,16-22,24

The same applies with other list formats, like {1 2 3} and (1; 2; 3) etc. In doubt? Ask!

Answer 1

Python 2, 132 115 bytes

^{-17 bytes thanks to Leaky Nun}

x,n=input()
o=[]
i=1
while x:
 t=x[0]
 while[t+i]==x[i:i+1]:i+=1
 o+=[[t,'-',t+i-1],x[:i]][i<n];x=x[i:];i=1
print o

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Answer 2

Jelly,  26 25  23 bytes

-2 bytes thanks to Erik the Outgolfer (by bringing the if statement into the main link)

Ḣ;Ṫj”-
IỊ¬1;œṗ⁸¹ÇL<¥?€F

A dyadic link returning a list in the [3,5,"-",8,10,"-",12,14,16,"-",22,24] format.

Try it online! (footer separates with spaces, to print the data example format).

How?

Ḣ;Ṫj”- - Link 1, format a run: list R
Ḣ      -     head
  Ṫ    -     tail
 ;     -     concatenate
    ”- -     literal '-'
   j   -     join

IỊ¬1;œṗ⁸¹ÇL<¥?€F - Main link: list L, number N
I                - incremental differences
 Ị               - insignificant? (<=1)
  ¬              - not
   1;            - prepend a 1
       ⁸         - chain's left argument, L
     œṗ          - partition (L) at truthy indexes
              €  - for €ach row, R, in L:
             ?   -   if:
            ¥    -   condition: last two links as a dyad:
          L      -     length of R
           <     -     is less than N?
        ¹        -   then: identity - do nothing, yields R
         Ç       -   else: call the last link (1) as a monad with argument  R
               F - flatten into a single list

Answer 3

Pyth, 23 bytes

sm?<ldvzd[hd\-ed).ga=hZ

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How it works

sm?<ldvzd[hd\-ed).ga=hZkQ

                        Q    autoinitialized to eval(input())
                 .g          group by k ↦
                    =hZ          Z += 1, returning new value (Z is autoinitialized to 0)
                   a   k         absolute difference with k
 m                           map d ↦
  ?                              if
    ld                               length of d
   <  vz                             less than eval(z) (z is autoinitialized to input())
        d                        then d
         [hd\-ed)                else [d[0], '-', d[-1]]
s                            concatenate

Answer 4

Japt, 24 bytes

óÈÄ¥Yîl ¨V?Zv +'-+Zo :Z

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Explanation

óÈ   Ä ¥ YÃ ®   l ¨ V?Zv +'-+Zo :Z
óXY{X+1==Y} mZ{Zl >=V?Zv +'-+Zo :Z}   Ungolfed
                                      Implicit: U = input array, V = input integer
óXY{      }                           Group U into runs such that for each pair X, Y:
    X+1==Y                              Y is exactly 1 more than X.
            mZ{                   }   Map each run Z to:
               Zl >=V?                  If Z has at least V items:
                      Zv     Zo           Z.unshift() and Z.pop() (the first and last items)
                         +'-+             joined with a hyphen.
                                :       Otherwise:
                                 Z        just Z.
                                      Implicit: output result of last expression

Answer 5

Retina, 101 bytes

\d+
$*
\b(1+) (?=1\1\b)
$1X
T`X` `\b((X)|1)+\b(?=.*¶(?<-2>1)+(?(2)(?!))11)
T`X`-
-1+(?=-)|¶1+

1+
$.&

Try it online! Takes the space-separated list L on the first line and the integer N on the second line. Explanation: The first stage converts the input to unary. The second stage changes the space between consecutive integers to an X. The third stage looks for runs of consecutive integers whose length is less than N and changes their Xs back to spaces. The fourth stage changes the Xs to - (this was 3 bytes shorter than using -s in the first place.) The fifth stage deletes all integers still left in the middle of a run, as well as N, while the final stage converts back to decimal.

Answer 6

Mathematica, 128 bytes

(s=#2;t=r=1;While[t<Length@s,If[s[[t+1]]-s[[t]]==1,r++,r=1];If[r==#,s[[t-#+3;;t]]="-";r--];t++];s//.{b___,a_,a_,c___}:>{b,a,c})&

input

[3,{3,5,6,7,8,10,11,12,14,16,17,18,19,20,21,22,24}]

output

{3, 5, "-", 8, 10, "-", 12, 14, 16, "-", 22, 24}

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Answer 7

APL, 38 bytes

{∊⍺{⍺>≢⍵:⍵⋄2⌽'-',2↑¯1⌽⍵}¨⍵⊂⍨1,1≠2-⍨/⍵}

Answer 8

PHP 7, 137 136 134 117 110 108 bytes

for($a=$argv,$i=2;$n=$a[$i++];$k<$a[1]||array_splice($a,$i,$k-2,"-"))for($k=print"$n ";$a[$i+$k]-++$k==$n;);

Takes L from first argument, list elements after that. Run with -nr or try it online.

Replace $L=($a=$argv) with $a=$argv,$L= (+1 byte) for PHP<7.

breakdown

for($a=$argv,$i=2;              # import input
    $n=$a[$i++];                # loop $n through list elements
    $k<$a[1]||                      # 3. if streak length ($k) is >=L ($a[1])
        array_splice($a,$i,$k-2,"-")    # then replace with "-"
)
for($k=print"$n ";                  # 1. print element and space
    $a[$i+$k]-++$k==$n;);           # 2. find consecutive numbers

Answer 9

Ruby, 68 bytes

->n,l{l.slice_when{|x,y|x<y-1}.map{|x|x[n-1]?x.minmax.uniq*?-:x}*?,}

Returns a string like for example 3,5-8,10-12,14,16-22,24.

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Answer 10

J, 40 bytes

;@((](,_,{:)/)^:(<:#)&.>]<;.1~1,1<}.-}:)

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Uses _ instead of -.

Explanation

;@((](,_,{:)/)^:(<:#)&.>]<;.1~1,1<}.-}:)  Input: integer N (LHS), array L (RHS)
                                  }.      Behead L
                                     }:   Curtail L
                                    -     Subtract elementwise to get the increments
                                1<        Test if greater than 1
                              1,          Prepend a 1
                        ]                 Get L
                         <;.1~            Partition L into boxes using the previous array
                     & >                  Operate on each box (partition) with N
              ^:                            If
                   #                          The length of the partition
                 <:                           Is greater than or equal to N
   (](     )/)                                Reduce (right-to-left) it using
         {:                                     Tail
       _,                                       Prepend _
      ,                                         Append to LHS
                     &.>                    Box the result
;@                                        Raze - join the contents in each box

Answer 11

Vyxal, 23 bytes

D0J¨pε⁽ṅḋ•ƛL⁰≥[₍ht\-Y}f

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23 bytes seems to be the standard here, so I figured I'd add my own 23 byter. Takes List then N.

Explained

D0J¨pε⁽ṅḋ•ƛL⁰≥[₍ht\-Y}f
D0J                     # Push 3 copies of the input and append 0 to the last
   ¨pε                  # get the absolute difference of each overlapping pair
      ⁽ṅḋ               # group by diff <= 1
         •              # mold to shape
          ƛ          }  # to each consecutive run:
           L⁰≥[         #   if the length is >= N
               ₍ht\-Y   #     [head, "-", tail]
                      f # flatten that

Answer 12

Jelly, 39 37 36 bytes

IỊṣ0;€1ṁ@
ÇL€<Ɠ¬TịÇḊ€Ṗ€F;€”-FyµŒgQ€F

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Takes the array via arguments, and the integer via STDIN. The TIO link uses the footer ÇG so the output is space-separated.

How? (Array: a, Integer: n)

(`f`)
IỊṣ0;€1ṁ@
I          Deltas of `a`
 Ị         Insignificant (x -> abs(x)<=1) applied to each element
  ṣ0       Split at occurrences of `0`.
    ;€1    Append `1` to each element
       ṁ@  `a` shaped like that
ÇL€<Ɠ¬TịÇḊ€Ṗ€F;€”-FyµŒgQ€F
Ç                            `f`
 L€                          Length of each element
   <Ɠ                        x -> x < n applied to each element
     ¬                       Logical not of each element (because Jelly doesn't have <= nor >= atoms)
      T                      Nonzero indexes
       ịÇ                    Index `f` at those indexes
         Ḋ€Ṗ€                x -> x[1:-1] applied to each element
             F               Flatten
              ;€”-           Append a hyphen to each element
                  F          Flatten
                   y         Translate (replaces all elements to be deleted with a hyphen)
                    µ        Start a new monadic link
                     Œg      Group runs of equal elements
                       Q€    Uniquify each element (make runs of hyphens one hypen)
                         F   Flatten, yet again.

I'm guessing I fell ... flat on this one.

Answer 13

JavaScript (ES6), 126 119 bytes

(e,c)=>{for(i=0,R='';i<e.length;R+=(R&&',')+(u-m>=c?m+'-'+--u:e.slice(z,i))){m=u=e[i],z=i;while(e[++i]==++u);}return R}

An anonymous function. Takes input in the order Array L, Integer N and returns the result as a comma-separated string.

M=(e,c)=>{for(i=0,R='';i<e.length;R+=(R&&',')+(u-m>=c?m+'-'+--u:e.slice(z,i))){m=u=e[i],z=i;while(e[++i]==++u);}return R}
console.log(M(eval(prompt()),+prompt()))

Answer 14

Dyalog APL v16.0, 82 80 78 76 75 65 62 bytes

{S/⍨1,⍨2≠/S←'-'@(⍸⊃∨/(-0,⍳⍺-3)⌽¨⊂(⍴⍵)↑∧/¨(⍺-1),/¯1⌽1=-2-/⍵)⊢⍵}

Wow, this is ... bad. There's probably a much much shorter solution with stencil.

Try it online!

Golfing suggestions welcome!