31
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We'll call the consecutive distance rating of an integer sequence the sum of the distances between consecutive integers. Consider 2 9 3 6 8 1.

2 9 3 6 8 1
<----5---->
<-2->
  <--3-->

\$2\$ and \$1\$ are consecutive integers, and their distance apart in the sequence is \$5\$.
\$2\$ and \$3\$ are consecutive integers, and their distance apart in the sequence is \$2\$.
\$9\$ and \$8\$ are consecutive integers, and their distance apart in the sequence is \$3\$.
The consecutive distance rating is the sum of these distances: \$10\$.

Challenge

Given a possibly empty list of positive, unique integers, find its consecutive distance rating.

Format

You must accept a list of integers and output an integer in any reasonable format.

Rules

Test cases

[] -> 0
[33] -> 0
[65 57 78 32 81 19 50 24 85 3 97 43 10 73] -> 0
[1 2] -> 1
[2 1] -> 1
[1 2 3] -> 2
[1 3 2] -> 3
[31 63 53 56 96 62 73 25 54 55 64] -> 26
[54 64 52 39 36 98 32 87 95 12 40 79 41 13 53 35 48 42 33 75] -> 67
[94 66 18 57 58 54 93 53 19 16 55 22 51 8 67 20 17 56 21 59] -> 107
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7
  • \$\begingroup\$ What about something like 3 2 2? \$\endgroup\$
    – emanresu A
    Jun 15, 2021 at 9:20
  • 1
    \$\begingroup\$ @Ausername The input will only consist of unique integers. \$\endgroup\$
    – chunes
    Jun 15, 2021 at 9:21
  • \$\begingroup\$ will the input contain only positive integers? \$\endgroup\$
    – Wasif
    Jun 15, 2021 at 9:29
  • 2
    \$\begingroup\$ @Wasif "Given a possibly empty list of positive, unique integers, find its consecutive distance rating." [my emphasis] \$\endgroup\$
    – Adám
    Jun 15, 2021 at 9:34
  • 1
    \$\begingroup\$ @Shaggy I'm going to say no, because the output format has been an integer from the beginning, and other languages may have had to compromise in bytes to satisfy that requirement. \$\endgroup\$
    – chunes
    Jun 15, 2021 at 23:49

23 Answers 23

Reset to default
10
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Python 2, 60 bytes

lambda l:sum(abs(l.index(x)-l.index(x+(x+1in l)))for x in l)

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3
  • 1
    \$\begingroup\$ There is a way of saving a few bytes by getting rid of the if ... \$\endgroup\$
    – ovs
    Jun 15, 2021 at 10:12
  • \$\begingroup\$ @ovs Ah, right, nice one \$\endgroup\$
    – xnor
    Jun 15, 2021 at 10:19
  • \$\begingroup\$ 1 in -> 1in \$\endgroup\$
    – user100690
    Jun 15, 2021 at 10:19
8
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J, 22 bytes

[:+/&,|@(#\-/#\)*1=-/~

Try it online!

Consider f 2 9 3 6 8 1:

  • -/~ Table of differences:

    0 _7 _1 _4 _6 1
7  0  6  3  1 8
1 _6  0 _3 _5 2
4 _3  3  0 _2 5
6 _1  5  2  0 7
_1 _8 _2 _5 _7 0

  • 1= Where are they 1?

    0 0 0 0 0 1
0 0 0 0 1 0
1 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0

  • |@(#\-/#\) Table of absolute value of index differences:

    0 1 2 3 4 5
1 0 1 2 3 4
2 1 0 1 2 3
3 2 1 0 1 2
4 3 2 1 0 1
5 4 3 2 1 0

  • * Elementwise product of those tables:

    0 0 0 0 0 5
0 0 0 0 3 0
2 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0

  • [:+/&, Sum of that table flattened:

    10
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8
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Jelly, 8 bytes

iⱮ‘aạ¥JS

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-1 byte thanks to Unrelated String!

How it Works

iⱮ‘aạ¥JS - Main link. Takes a list L on the left
  ‘      - Increment all elements in L
 Ɱ       - Over each incremented element:
i        -   Get its 1-based index in L, or 0 if not present
           Call this list of indices I
      J  - Indices of L; [1, 2, 3, ..., len(L)]
     ¥   - Last 2 links as a dyad f(I, J):
    ạ    -   Absolute difference, element wise, between I and J
   a     -   And; Where there are non-zeroes in I, replace the value with corresponding difference
       S - Sum
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2
5
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05AB1E, 8 bytes

ãʒÆ}kÆÄO

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ã          # all pairs of integers from the input
 ʒ }       # keep those where the following results in 1:
  Æ        #   reduce by subtraction
    k      # for each integer in each pair get the index in the input list
     Æ     # reduce each pair of indices by subtraction
      Ä    # take absolute value
       O   # sum all distances
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5
\$\begingroup\$

Red, 88 77 bytes

func[b][s: 0 forall b[s: s + absolute offset? b any[find head b 1 + b/1 b]]s]

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Uses @xnor's method - don't forget to upvote it!

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3
  • \$\begingroup\$ Is the /: an assign at index? \$\endgroup\$
    – Razetime
    Jun 15, 2021 at 10:49
  • \$\begingroup\$ @Razetime No, it's just indexing using the word (think variable) i. You can't just use b/i because it would mean select b 'i and the block b doesn't have a word i inside itself. \$\endgroup\$
    – Galen Ivanov
    Jun 15, 2021 at 10:53
  • \$\begingroup\$ @Razetime Assign at index is b/1: value. If the index is referred to by a word idx, we need to write b/:idx: value. \$\endgroup\$
    – Galen Ivanov
    Jun 15, 2021 at 11:49
5
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R, 47 bytes

sum(abs(seq(a<-scan())-(b=match(a+1,a,0)))*!!b)

Try it online! with test wrapper stolen from pajonk's answer stolen from Kirill L's answer

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2
  • \$\begingroup\$ The test wrapper is actually by @Kirill ;-) \$\endgroup\$
    – pajonk
    Jun 15, 2021 at 16:01
  • \$\begingroup\$ Then why not merge our solutions? 45 bytes \$\endgroup\$
    – pajonk
    Jun 15, 2021 at 16:09
5
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R, 45 bytes

sum(abs(seq(a<-scan())-match(a+1,a)),na.rm=T)

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Merged solutions: mine and @Dominic's.

Previous solution:

R, 48 bytes

x=scan();`*`=match;sum(abs(x*x-(x+1)*x),na.rm=T)

Try it online!

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3
  • \$\begingroup\$ This can be 48 as a full program, making it the same length as my answer. However, now when there's \(x) syntax in R 4.1, but not available on TIO, I'm not sure anymore, which format to prefer... \$\endgroup\$
    – Kirill L.
    Jun 15, 2021 at 14:04
  • 1
    \$\begingroup\$ @Kirill I'll change it to the scan version, so the answers are comparable. Which format to prefer? I mainly post functions because I think it's easier to read that way and I mostly care about the internal algorithm not the wrapper or technicalities. \$\endgroup\$
    – pajonk
    Jun 15, 2021 at 14:46
  • 1
    \$\begingroup\$ I couldn't resist joining the competition... \$\endgroup\$
    – Dominic van Essen
    Jun 15, 2021 at 15:24
4
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JavaScript (Node.js), 60 bytes

n=>n.map((e,i)=>z+=(j=n.indexOf(e+1))+1?j>i?j-i:i-j:0,z=0)|z

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2
  • 1
    \$\begingroup\$ (j=n.indexOf(e+1))+1 ~> ~(j=n.indexOf(e+1)) \$\endgroup\$
    – Arnauld
    Jun 15, 2021 at 10:24
  • \$\begingroup\$ (j=n.indexOf(e+1))<i?~j&&i-j:j-i is better still. \$\endgroup\$
    – Neil
    Jun 15, 2021 at 18:37
4
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R, 48 bytes

`?`=diff;sum(abs(?o<-order(x<-scan()))*!1<?x[o])

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1
  • \$\begingroup\$ Wow, that's impressive! Great use of order! \$\endgroup\$
    – pajonk
    Jun 15, 2021 at 14:42
4
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Ruby, 50 48 bytes

->a{r=-1;a.sum{((a.index(a[r+=1]+1)||r)-r).abs}}

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Thanks @ovs for -2 bytes.

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0
4
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Brachylog, 15 bytes

sᶠ{↻h₂-ȧ1&b}ˢcl

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sᶠ                 Find every substring of the input.
  {        }ˢ      Keep only those for which
       ȧ           the absolute value of
      -            the difference of
   ↻h₂             the first and last elements
        1          is 1,
         &b        and remove their first elements.
             c     Concatenate them,
              l    and output their combined length.
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3
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Charcoal, 15 bytes

Ｉ↨¹Ｅθ↔↨¹⁻⌕Ａθ⊕ικ

Try it online! Link is to verbose version of code. Explanation:

    θ           Input array
   Ｅ            Map over elements
             ι  Current value
            ⊕   Incremented
         ⌕Ａ     Find all (i.e. at most one) matches
           θ    In input array
        ⁻       Vectorised subtract
              κ Current index
      ↨¹        Take the sum
     ↔          Take the absolute value
 ↨¹             Take the sum
Ｉ               Cast to string
                Implicitly print

Note that base 1 conversion is used to sum as this works on empty lists.

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3
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Red, 129 bytes

func[a][s: 0
b: sort copy a
while[b/2][s: s + either b/2 - b/1 = 1[absolute(index? find a b/2)- index? find a b/1][0]b: next b]s]

Try it online!

sorts the array, and adds the index difference to s if pair of elements has an absolute difference of 1.

Not sure how to shorten the reuse of index? find, but I think it can be shortened. abs alias for absolute is not available, even in version 0.6.4.

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3
\$\begingroup\$

Excel, 52 bytes

=LET(x,A:A,SUM(IFERROR(ABS(ROW(x)-XMATCH(x-1,x)),)))

Input numbers in column A. For each number x in column A look for x-1. If found add the difference in rows between x and x-1.

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3
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Vyxal s, 13 bytes

›vḟv›ƛ&›[¥ε|0

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A port of the jelly answer, and I think the best I'm gonna get.

Attempted port of Jonas, 21 bytes

Ẋƛ÷ε1=;?Lɾƛ?Lɾ-ȧ;f*∑½

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Old version, 23 bytes

L²(⁰Ẋni÷ε1=[n6ḋ÷ε&+])¥½

Try it Online! I still think there's a bit more I can do.

Older version, 28 bytes

L(x|L(←x i⁰niε1=[n←xε&+]))¥½

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This. Is. Horrible.

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1
  • \$\begingroup\$ Porting jelly somehow could save bytes \$\endgroup\$
    – lyxal
    Jun 15, 2021 at 9:41
3
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MATL, 9 bytes

tQ!=&f-|s

Try it online! Or verify all test cases.

How it works

Consider input [2 9 3 6 8 1] as an example.

tQ   % Implicit input. Duplicate, add 1; element-wise
     % STACK: [2 9 3 6 8 1], [3 10 4 7 9 2]
!    % Transpose
     % STACK: [2 9 3 6 8 1], [3; 10; 4; 7; 9; 2]
=    % Test for equality (element-wise with broadcast)
     % STACK: [0 0 1 0 0 0;
               0 0 0 0 0 0;
               0 0 0 0 0 0;
               0 0 0 0 0 0;
               0 1 0 0 0 0;
               1 0 0 0 0 0]
&f   % Two-output find: row and column indices of nonzeros
     % STACK: [6; 5; 1], [1; 2; 3]
-|   % Minus, absolute value (element-wise)
     % STACK: [5; 3; 2]
s    % Sum. Implicit display
     % STACK: 10
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3
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Japt -mx, 16 11 10 bytes

Inspired by xnor's solution.

VaWbU+WøUÄ

Try it

VaWbU+WøUÄ     :Implicit map of each U at 0-based index V in input array W
Va             :Absolute difference of V and
  Wb           :  Index in W of
    U+Wø       :    U + Does W contain
        UÄ     :      U+1
               :Implicit output of sum of resulting array
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3
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Julia 1.0, 63 55 bytes

Inspired by the answer of xnor

Improved version, thanks @MarcMush:

!x=sum(abs,[(I=indexin)(i,x)-I(i+(i+1∈x),x) for i=x])

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First version:

!x=sum(abs.([indexin(i,x)-indexin(i+(i+1 in x),x) for i in x]))

Try it online!

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2
2
\$\begingroup\$

Python 3, 91 bytes

def f(l):c=sorted(l);return sum(abs(l.index(x)-l.index(y))for x,y in zip(c,c[1:])if y-x==1)

Try it online!

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1
  • 1
    \$\begingroup\$ -1 by switching to Python 3.8. \$\endgroup\$
    – user100690
    Jun 15, 2021 at 10:27
2
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Wolfram Language (Mathematica), 42 bytes

Tr@Abs[s@@@Outer[s=#-#2&,#,#]~Position~1]&

Try it online!

Essentially a port of Jonah's J answer.

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2
\$\begingroup\$

Japt -x, 20 17 12 bytes

ã_ÎaZo)¥1©ZÊ

Try it

  • saved 5 thanks to @Shaggy!
    ã_  - subsections passed trough
    ÎaZo) * abs difference of 1st
            and last element(then removed)
    ¥1    * == 1?
    ©ZÊ   * take lenght/else 0
    -x flag to sum
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5
  • \$\begingroup\$ 15 bytes, or 14 bytes with the -x flag. \$\endgroup\$
    – Shaggy
    Jun 16, 2021 at 8:57
  • \$\begingroup\$ 13 bytes \$\endgroup\$
    – Shaggy
    Jun 16, 2021 at 9:03
  • \$\begingroup\$ 12 bytes; always forget A.ã() can take a function. I hate that I've just outgolfed myself! \$\endgroup\$
    – Shaggy
    Jun 16, 2021 at 9:06
  • \$\begingroup\$ @Shaggy great! Didn't notice A.o removed the element, that's why it gave me wrong result! \$\endgroup\$
    – AZTECCO
    Jun 16, 2021 at 9:33
  • 1
    \$\begingroup\$ A.o(), without an argument, is the same as .pop() in JS. \$\endgroup\$
    – Shaggy
    Jun 16, 2021 at 17:35
2
\$\begingroup\$

JavaScript (V8), 54 bytes

$=([o,...a])=>o?a.indexOf(o+1)+a.indexOf(o-1)+2+$(a):0

Try it online!

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1
\$\begingroup\$

Thunno, \$ 24 \log_{256}(96) \approx \$ 19.75 bytes

2zPgAu-1=keez0AhEeAuADES

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Explanation

2zP       # All pairs in the input
   g      # Filtered by:
    Au-   #  Does the difference
       1= #  Equal 1?
k         # End filter
 e        # Map over the list:
  e       #  Map over the pair
   z0Ah   #   Get the index in the input
       E  # End both maps
e         # Map over this list:
 AuAD     #  Get the absolute difference
     E    # End map
      S   # Sum this list
          # Implicit output
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\$\endgroup\$

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