Sort a difference list

Question

The difference list of a list of integers is the list differences of consecutive members.

For example the difference list of

1, 3, 2 ,4

is

2, -1, 2

Your task is to take as input a difference list and output what the difference list would look like if the original list were sorted.

For example the difference list

2, 1, -2, -1

Might represent a list

2 4 5 3 2

Which when sorted is

2 2 3 4 5

Which has a difference list of

0 1 1 1

---

This is code-golf so answers will be scored in bytes with less bytes being better.

Answer 1

05AB1E, 4 bytes

.¥{¥

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Explanation

.¥{¥
.¥   # Undelta the input list
  {  # Sort it
   ¥ # And get the deltas

Answer 2

Python 3 with Numpy, 56 54 53 bytes

2 bytes off thanks to @Artyer (Numpy's sort instead of standard sorted). 1 byte off thanks to @notjagan (moving 0 into cumsum)

lambda x:diff(sort(cumsum([0]+x)))
from numpy import*

The code defines an anonymous function that inputs a list or a Numpy array and outputs a Numpy array.

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Answer 3

Mathematica, 40 bytes

Differences@Sort@Accumulate@Join[{1},#]&

Answer 4

JavaScript (ES6), 57 56 bytes

Saved 1 byte thanks to @ETHproductions

a=>a.map(n=>t-=n,p=t=0).sort((a,b)=>b-a).map(n=>p-(p=n))

Demo

let f =

a=>a.map(n=>t-=n,p=t=0).sort((a,b)=>b-a).map(n=>p-(p=n))

console.log(JSON.stringify(f([2, 1, -2, -1])))

Answer 5

Jelly, 6 bytes

0;+\ṢI

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Answer 6

Husk, 4 bytes

Ẋ-O∫

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Explaination

      -- implicit input, e.g                               [2,1,-2,-1]
   ∫  -- cumulative sum                                    [0,2,3,1,0]
  O   -- sort                                              [0,0,1,2,3]
Ẋ     -- apply function to all adjacent pairs in the list  [(0,0),(0,1),(1,2),(2,3)]
 -    --   subtract                                        [0,1,1,1]

Answer 7

Pyth, 9 bytes

-1 byte thanks to @EriktheOutgolfer.

.+S+0sM._

Test Suite.

Pyth, 10 bytes

.+S.u+YNQ0

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Answer 8

Java 8, 123 bytes

The standard solution: cumulative sum input, sort, then diff. No substantial implementation tricks either.

l->{int s=l.length,d[]=new int[s+1],i=0;while(i<s)d[i+1]=d[i]+l[i++];for(java.util.Arrays.sort(d);i-->0;)l[i]=d[i+1]-d[i];}

Cast to Consumer<int[]>. Output is mutated input.

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Ungolfed lambda

l -> {
    int
        s = l.length,
        d[] = new int[s + 1],
        i = 0
    ;
    while (i < s)
        d[i + 1] = d[i] + l[i++];
    for (java.util.Arrays.sort(d); i-- > 0; )
        l[i] = d[i + 1] - d[i];
}

Acknowledgments

• -3 bytes thanks to Olivier Grégoire, master of unholy autoincrementation
• -1 byte thanks to Nevay

Answer 9

R, 31 32 bytes

-4 bytes thanks to @user2390246 for diffinv

+5 bytes from Jarko for cat

cat(diff(sort(diffinv(scan()))))

Reads from stdin, writes to stdout. diffinv is an inverse of diff for a given starting value (0 by default). Since it's diffed again, it doesn't matter what that value is.

As pointed out by Jarko Dubbeldam, I needed to properly output the result, at the cost of five bytes. Alas.

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Answer 10

Brachylog, 15 bytes

a₀ᶠ+ᵐ,0o{s₂↔-}ᶠ

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Answer 11

Perl 6, 46 bytes

{[\+](0,|@_).sort.rotor(2=>-1).flat.map(*R-*)}

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Expanded:

{  # bare block lambda with implicit signature :(*@_)

  [\+](         # triangle reduce using &infix:«+»
    0,          # start with 0
    |@_         # Slip in the arguments from the outer block
  )             #                  (0, 2, 3, 1, 0)

  .sort         # sort the results (0,0,1,2,3)
  .rotor(2=>-1) # group in twos    ((0,0),(0,1),(1,2),(2,3))
  .flat         # flatten          (0,0,0,1,1,2,2,3)
  .map(*R-*)    # grab 2 values at a time, and subtract first from second
                # (0, 1, 1, 1)
}

Answer 12

Python 2, 83 bytes

l,r=input(),[1]
for i in l:r+=[r[-1]+i]
r.sort()
print[b-a for a,b in zip(r,r[1:])]

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Horrible solution.

Answer 13

VB.NET (.NET 4.5), 109 bytes

Sub A(n)
Dim c=n.count-1
For i=1To c
n(i)+=n(i-1)
Next
n.Sort()
For i=c To 1 Step-1
n(i)-=n(i-1)
Next
End Sub

A function that expects a list as input and modifies it directly. The original parameter can then be used for output

1. Recreates an original list by adding forwards through the list (assumes an implicit 0 as the first element)
2. Sorts the original list
3. Gets the differences by going backwards (so I don't need to keep track of a different list) (the implicit first element of 0 means the first difference is the same as the smallest element)

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Answer 14

Haskell, 74 bytes

import Data.List
g=sort.scanl(+)0
h l|k<-g l=map(\(x,y)->x-y)$zip(tail$k)k

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Straightforward.

Answer 15

TI-Basic (TI-84 Plus CE), 23 bytes

Prompt X
augment({0},cumSum(LX→X
SortA(LX
ΔList(LX

Prompts for user input. The list must be input with a leading {, with numbers separated by ,, and with an optional trailing }.

TI-Basic is a tokenized language; ΔList( and cumSum( are two-byte tokens, all other tokens used are one byte each.

Example run (with NAME as the program name and {4,-2,7,-4,0} as the input):

prgmNAME
X=?{4,-2,7,-4,0}
               {2 2 1 0 4}

Explanation:

Prompt X                  # 3 bytes, get list input, store in LX
augment({0},cumSum(LX→X   # 12 bytes, 
          # store the list ({0} prepended to the cumulative sum of LX) to LX
SortA(LX                  # 4 bytes, sort LX ascending
ΔList(LX                  # 4 bytes, implicitly print the difference list of LX

Answer 16

C++ (gcc), 136 bytes

As unnamed generic lambda, assuming input to be like std::list and returning via reference parameter.

[](auto&L){auto r=L.begin(),l=L.insert(r,0);while(r!=L.end())*r+++=*l++;for(L.sort(),l=r=--L.end();--l!=L.begin();*r---=*l);L.erase(l);}

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Ungolfed:

[](auto&L){
 auto r=L.begin(),
      l=L.insert(r,0); //adds a zero right in front
 while(r!=L.end())
   *r++ += *l++;       //sum left to right
 for(
  L.sort(),            //sorting invalidates the iterators
  l=r=--L.end();       //so, reinit
  --l!=L.begin();      //decrement l beforehand 
  *r-- -= *l           //diff right to left
 );
 L.erase(l);           //l==L.begin(), so this removes the temporary 0
}

Answer 17

Pyth, 8 bytes

.+S+M.uP

Demonstration

.+S+M.uP
.+S+M.uPNQ    Implicit variables
     .u  Q    Apply the following function to the input repeatedly until it
              stops changing, then output the list of values, including the
              starting value.
       PN     Remove the last element. No-op if the list is empty.
   +M         Sum each list. This gives the cumulative sums in reverse order,
              including a 0 at the end for the empty list.
  S           Sort
.+            Deltas

Answer 18

TI-Basic, 20 bytes

cumSum(augment({0},Ans->L1
SortA(L1
ΔList(L1

Answer 19

Perl 5, 87 + 1 (-a) = 88 bytes

$a[0]=1;push@a,$a[-1]+$_ for@F;@a=sort{$a<=>$b}@a;print$a[0]-$_,$"while($_=shift@a)&&@a

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Answer 20

APL (Dyalog), 15 14 bytes

-1 byte thanks to ngn.

(¯2-/⍋⊃¨⊂)0,+\

+\ cumulative sum

0, prepend a zero

(…) apply the following tacit function on that:

 ⊂ enclose (so we can pick multiple items)

 ⍋⊃¨ let each of the indices that would sort the argument pick from that

 ¯2-/ reversed pairwise difference

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---

Original solution found by the Code Golf Hackathon participants at the Dyalog '17 User Meeting:

¯2-/l[⍋l←+\0,⎕]

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⎕ prompt for input

0, prepend a zero

+\ cumulative sum

l← store as l

⍋ find the indices that will sort l

l[…] use that to index into l

¯2-/ reversed pairwise difference

Answer 21

MATL, 6 bytes

0hYsSd

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0       # push 0
 h      # horizontal concatenate with implicit input
  Ys    # cumulative sum
    S   # sort
     d  # diff (implicit output)

Answer 22

Gaia, 7 bytes

1¤++⊣ȯọ

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Explanation

1¤+      Prepend 1
   +⊣    Cumulative sums
     ȯ   Sort
      ọ  Deltas

Answer 23

k, 16 bytes

1_-':{x@<x}@+\0,

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              0, /prepend a 0
            +\   /cumulative sum
     {x@<x}@     /sort
1_-':            /differences list

Answer 24

Japt, 10 bytes

iT å+ n än

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Answer 25

Röda, 42 bytes

{i=0{[0];[i]if i+=_}|sort|slide 2|[_2-_1]}

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This is similar to the Perl 6 answer. .sort is |sort, .rotor(2=>-1).flat is |slide 2 and .map(*R-*) is |[_2-_1].

Explanation:

{
  i=0 /* initialize variable i */
  /* the following block recreates the original list from differences: */
  {
    [0];       /* push 0 to the stream */
    [i]if i+=_ /* add every number in the stream to i and push i back */
  }|
  sort|    /* sort the numbers */
  slide 2| /* for values i1, i2, i3, ... in the stream
              push pairs i1, i2, i2, i3, ... */
  [_2-_1]  /* calculate difference of numbers in each pair in the stream */
}

The statement [i]if i+=_ is equivalent to

for sfv do
  if i += sfv do
    push(i)
  done
done

The += operator does not push values to the stream, so it is truthy. I could also have used some kind of block (eg. {|j|i+=j;[i]}_) to tie the addition and pushing statements together, but if is shorter.

Answer 26

Julia 0.6.0 (34 bytes)

Pretty much a copy of what has been done in R and Python 3

x->diff(sort(cumsum(vcat([0],x))))

Answer 27

J, 10 bytes

/:~&.(+/\)

explanation

"sort under scan sum": In J, the Under conjunction &. applies the transformation to its right to the input, then applies the verb to its left (in this case sort /:~) and then does the reverse transformation. That is, J understands how to invert a scan sum, which is exactly what's needed here: the successive differences are the input that, when scan-summed, will produce that scan-sum.

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