Ways to add 1 to lists of lists

Question

Adapted from this StackOverflow question

In this challenge you will take a list of lists of integers, e.g.

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

And an additional single integer (e.g. n = 7). If you were to add n to the front of one of the lists in A there would be as many ways to do that as there are lists in A. In this example 4:

A' = [[7,1,2],[3,4],[5],[]]
A' = [[1,2],[7,3,4],[5],[]]
A' = [[1,2],[3,4],[7,5],[]]
A' = [[1,2],[3,4],[5],[7]]

In this challenge you will output all possible ways to do this, in the order of how early n is inserted. So for the example the output is just:

[ [[7,1,2],[3,4],[5],[]]
, [[1,2],[7,3,4],[5],[]]
, [[1,2],[3,4],[7,5],[]]
, [[1,2],[3,4],[5],[7]]
]

This is codegolf so answer answers will be scored in bytes with fewer bytes being better.

Test cases

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

Answer 1

Rust, 185 184 bytes

My first ever code golf submission

fn r(a:Vec<Vec<u8>>,n:u8){let mut x:Vec<Vec<Vec<u8>>>=Vec::new();for i in 0..a.len(){let mut b=a.clone();let c=&vec![n];b[i].splice(0..0,c.iter().cloned());x.push(b);}print!("{:?}",x)}

Explanation

This code copies the list for every sublist, then appends the number to the front of it, adds it to a final list which it then prints. We can also get away with returning the print statement to save a ;!

Try it online!

Edit: Kevin Cruijssen pointed out a space in my print.

Edit: @Bubbler showed me what it's like to truly golf in Rust and got it down to 90 bytes! I won't be updating the byte count in the header since I don't believe that his edits counts as "edits" rather an entire new submission and it doesn't reflect my work.

|a:&mut[Vec<_>],n|{for i in 0..a.len(){a[i].insert(0,n);print!("{:?}",a);a[i].remove(0);}}

Answer 2

05AB1E, 6 bytes

UεXšNǝ

Try it online or verify all test cases.

Explanation:

U       # Pop the store the first (implicit) input-integer in variable `X`
 ε      # Map over the second (implicit) input-list of lists:
  Xš    #  Prepend `X` in front of the current part
    Nǝ  #  And replace the item at the current map-index in the second (implicit)
        #  input-list with this modified part
        # (after which the result is output implicitly)

Answer 3

Haskell, 37 bytes

This is Willem Van Onsem's SO answer with a couple of trivial golfs.

n!(a:b)=((n:a):b):map(a:)(n!b)
_!_=[]

Try it online!

Answer 4

APL+WIN, 36 bytes

Prompts for the list of lists as a nested vector then the integer to be added. Index origin = 0

m←((⍴n)*2)⍴n←,⎕⋄m[(1+⍴n)×⍳⍴n]←⎕,¨n⋄m

Try it online! Thanks to Dyalog Classic

Answer 5

J, ^{22 17 21} 20 bytes

(<@#"0~[:=#\),&.>"1]

Try it online!

Could save 4 bytes if we can assume each element of the input is unique.

Consider 7 f 1 2 3; 1 2; 4:

• [:=#\ Create an identity matrix whose sides equal our list length:

  1 0 0
  0 1 0
  0 0 1
  

• <@#"0~ Use that as a mask to copy our new element, and box each result. Zeros become empty boxes:

  ┌─┬─┬─┐
  │7│ │ │
  ├─┼─┼─┤
  │ │7│ │
  ├─┼─┼─┤
  │ │ │7│
  └─┴─┴─┘
  

• ,&.>"1] For each row, join elementwise to the original input:

  ┌───────┬─────┬───┐
  │7 1 2 3│1 2  │4  │
  ├───────┼─────┼───┤
  │1 2 3  │7 1 2│4  │
  ├───────┼─────┼───┤
  │1 2 3  │1 2  │7 4│
  └───────┴─────┴───┘
  

Answer 6

JavaScript (Node.js), 41 bytes

t=>a=>a.map(u=>a.map(v=>u==v?[t,...u]:v))

Try it online!

Answer 7

Wolfram Language (Mathematica), 31 bytes

(i=1;x##~Insert~{i++,1})/@#&

Try it online!

Answer 8

Vyxal, 54 bits^v2, 6.75 bytes

vJ¨2¹^Ȧ

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Bitstring:

100010101100011111000100110100010110110111100010110010

Answer 9

Uiua, 13 12 bytes

≡⍜(⇌⊔⊡)⊂⇡⊃⧻¤

Try it!

-1 thanks to Bubbler

≡⍜(⇌⊔⊡)⊂⇡⊃⧻¤
         ⊃⧻¤  # fix the array and get its length
        ⇡     # range (of length)
≡⍜(⇌⊔⊡)       # change each unboxed reversed row by...
       ⊂      # join (the integer input)
              # then unreverse, re-box, and put each item back

Answer 10

Retina, 33 bytes

L$`\[+(?=(.+),(.+))
$>`$2,$1
,]
]

Try it online! Takes n as the second argument. Explanation:

L$`\[+(?=(.+),(.+))

Find the beginning of each list.

$>`$2,$1

For each occurrence, output the list with n moved to that position.

,]
]

Fix up the list if it was originally empty.

Answer 11

Python 3, 50 bytes

lambda a,L:[[[a][:l is k]+l for l in L]for k in L]

Try it online!

Will not work on lists with multiple references to the same object. I..e.: L = [[10],[10]] fine (same value but different objects) but L = 2*[[10]] fail (twice the same object)

Answer 12

R, 58 bytes

Or R>=4.1, 44 bytes by replacing two function occurrences with \s.

function(x,A)Map(function(i){A[[i]]=c(x,A[[i]]);A},seq(A))

Try it online!

Answer 13

Julia 1.0, 47 bytes

n\l=(k=keys(l)).|>i->k.|>j->[fill(n,i==j);l[j]]

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Answer 14

Ruby, 60 bytes

->n,l{(0...t=l.size).map{|i|[*l[0,i],[n]+l[i],*l[i+1...t]]}}

Try it online!

Answer 15

ayr, 15 bytes

Beats J by 5 bytes, so I'm happy :)

],`"\:]:#":i:&#

Explanation

           i:&#    Construct an NxN identity matrix (N = # lists)
        #":        Foreach num I in that, replace with left arg I times
      ]:           Convert that to a hook
    \:             For each item on the right and entirety of left
 ,`"               Concatenate left to back of right on an element x element basis
]                  Where the left is the list

Takes the number on the left and the list of lists on the right.

Answer 16

jq, 51 bytes

. as$a|[range(length)]|map(. as$i|$a|.[$i]|=[$n]+.)

Where $n is n.

Answer 17

Curry (PAKCS), 21 bytes

n!(a++b:c)=a++(n:b):c

Attempt This Online!

Answer 18

Charcoal, 16 bytes

ＩＥηＥη⎇⁼κμ⮌⊞Ｏ⮌λθλ

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

  η                 Input list of lists
 Ｅ                  Map over each list
    η               Input list of lists
   Ｅ                Map over each list
       κ            Outer index
      ⁼             Equal to
        μ           Inner index
     ⎇              If true then
             λ      Current list
            ⮌       Reversed
              θ     Input `n`
          ⊞Ｏ        Appended
         ⮌          Reversed
               λ    Otherwise current list
Ｉ                   Cast to string
                    Implicitly print

Charcoal's default output format might be a little tricky to read, so here's a 17-byte version with prettier output:

Ｅη⭆¹Ｅη⎇⁼κξ⮌⊞Ｏ⮌νθν

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

Answer 19

JavaScript (ES6), 44 bytes

Expects (n)(list_of_lists).

Version 1

n=>a=>a.map((v,i,[...b])=>(b[i]=[n,...v],b))

Try it online!

Version 2

n=>a=>a.map((_,i)=>a.map(v=>i--?v:[n,...v]))

Try it online!

Answer 20

Pari/GP, 48 bytes

(n,a)->matrix(#a,,i,j,concat([n][1..i==j],a[j]))

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Answer 21

Pip -xP, 13 bytes

aRA_(BPEb)MEa

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Explanation

The -x flag evaluates the inputs, meaning we can treat [[1];[2;3]] as a list rather than a string.

aRA_(BPEb)MEa
            a  First command-line input (the nested list)
          ME   Map the following function to that list, enumerated (first function arg
               is index, second function arg is sublist):
a               The whole list
 RA             Replace the element at index given by
   _            first function arg
    (    )      with
     B          second function arg
      PE        with the following value prepended:
        b       Second command-line input

The -P flag prints each sublist of the result, formatted as a list, on a separate line. Other formats that could work include -p and -S.

Answer 22

Ruby, 42 bytes

->n,l{i=-1;l.map{|w|z=*l;z[i+=1]=[n]+w;z}}

Try it online!

Answer 23

Burlesque, 24 bytes

1svsa.*zi{{1gv+]}x/ap}^m

Try it online!

I'm sure there's a better way...

1sv    # Save input as 1
sa.*   # Repeat arr len(arr) times
zi     # Zip with indices
{
 {
  1gv  # Get input (value1)
  +]   # Prepend
 }
 x/    # Reorder local stack
 ap    # Apply function to index (from zip)
}^m    # Push and map each

Answer 24

Stax, 10 bytes

ü↨○◘╞Q┐p☺Ç

Run and debug it