Find the J twin

Question

For a 2 dimensional array we will define the elements in either the first row or the last column to be the largest "J-Bracket" of the array. For example in the following array elements in the J-bracket are highlighted:

\$ \begin{bmatrix} \color{red}{\underline 1} & \color{red}{\underline 2} & \color{red}{\underline 4} & \color{red}{\underline 8} \\ 9 & 3 & 6 & \color{red}{\underline 7} \\ 3 & 3 & 2 & \color{red}{\underline 9} \end{bmatrix} \$

The J-bracket is given in order starting from the first element of the first row and going clockwise. The element that is in both the row and the column is not repeated. So for the above that is:

\$ \left[1, 2, 4, 8, 7, 9\right] \$

To get the next largest J-bracket is just remove the largest J-bracket from the array and take the largest J-bracket of the remainder:

\$ \begin{bmatrix} \color{lightgrey}{1} & \color{lightgrey}{2} & \color{lightgrey}{4} & \color{lightgrey}{8} \\ \color{red}{\underline 9} & \color{red}{\underline 3} & \color{red}{\underline 6} & \color{lightgrey}{7} \\ 3 & 3 & \color{red}{\underline 2} & \color{lightgrey}{9} \end{bmatrix} \$

and so on until every element is in exactly 1 J-bracket.

The set of J-brackets from an array is not necessarily unique. In fact if the matrix is non-square every matrix has a twin with the same J-bracket set.

\$ \begin{bmatrix} \color{red}{\underline 1} & \color{red}{\underline 2} & \color{red}{\underline 4} & \color{red}{\underline 8} \\ \color{green}9 & \color{green}3 & \color{green}6 & \color{red}{\underline 7} \\ \color{blue}{\overline 3} & \color{blue}{\overline 3} & \color{green}2 & \color{red}{\underline 9} \end{bmatrix} \cong \begin{bmatrix} \color{red}{\underline 1} & \color{red}{\underline 2} & \color{red}{\underline 4} \\ \color{green}{9} & \color{green}3 & \color{red}{\underline 8} \\ \color{blue}{\overline 3} & \color{green}6 & \color{red}{\underline 7} \\ \color{blue}{\overline 3} & \color{green}2 & \color{red}{\underline 9} \end{bmatrix} \$

This twin has the opposite dimensions and in the case of a square matrix it is its own twin.

Your task is to take a 2D array of positive integers and output its J-twin. You may take input and output in any standard format but the input format should be the same as the output.

This is code-golf so answers will be scored in bytes with fewer bytes being the goal.

Test cases

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

Answer 1

Jelly, 10 bytes

ZL_Lṙ@ŒdŒḍ

A monadic Link that accepts a rectangular list of lists and yields its J-bracket-twin as a list of lists.

Try it online! Or see the test-suite.

How?

ZL_Lṙ@ŒdŒḍ - Link: list of lists, A
Z          - transpose A
 L         - length of that -> columnCount(A)
   L       - length of A -> rowCount(A)
  _        - columnCount(A) subtract rowCount(A) = N
      Œd   - anti-diagonals of A
    ṙ@     - rotate this list of anti-diagonals left by N
        Œḍ - create a matrix with these anti-diagonals

Answer 2

K (ngn/k), 154 74 72 bytes

{$[1=&/#'1*:\x;:+x;[r:,/(*x),1_-1#'x;(,(#x)#r),+(+o(1_-1_'x)),,(#x)_r]]}

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Answer 3

Python 3, 111 bytes

def f(x):
 r,*a=x;k=len(a)
 if k*r[1:]:r+=map(list.pop,a);x=*zip(r[:k],*f(a)),r[k:]
 return[*map(list,zip(*x))]

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Answer 4

R, 76 bytes

\(m)array(unsplit(split(m,i),rev(t(i<-pmin(row(m),rev(col(m)))))),dim(t(i)))

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As in the other challenge, splits the array to find the J-brackets. Then unsplits and adds array structure to get to the J-twin as a matrix.

Longer explanation to follow.

Answer 5

J, 62 60 bytes

Quite messy. The logic is very similar to my Python answer. Porting the Jelly answer might be shorter.

[:|:*@#@,(({.,{:"1@}.)((}.~#),~({.~#),.])|:@$:@:(}:"1)@}.^:)

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Answer 6

PARI/GP, 59 bytes

m->matrix(#m,#m~,i,j,m[i-d=min(l=i+j-1,#m)-min(l,#m~),j+d])

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Shift the \$n\$-th anti-diagonal to lower left by \$\min(n,w)-\min(n,h)\$, where \$w\$ and \$h\$ are the number of columns and rows of the matrix respectively.

Answer 7

Python + NumPy, 80 bytes

def f(a):b=0*a.T;b[1:,:-1]=a.any()and f(a[1:,:-1]);b[b<1]=a[a>0];a[:]=0;return b

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Accepts and destroys (overwrites with zeros) a NumPy array. Relies on elements being positive as promised in OP.

Answer 8

Haskell, 132 121 bytes

-11 thanks to Wheat Wizard

t[h]=pure<$>h
t s@([x]:r)=[s>>=id]
t(h:r)|n<-t$init<$>r,(a,b)<-splitAt(length r)$h++(last<$>r)=zipWith(\i->(++[i]))b(a:n)

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If the input list has only one row, or only one column, transpose it. Otherwise, recurse on the "nub" that results from removing the J-bracket; then split the J-bracket into two parts, put one part atop the nub, and append the other part itemwise to the right side of the nub.

It's still pretty long, so I'm guessing there's a better way to do this...

Answer 9

Charcoal, 52 43 39 bytes

Ｆ⁺θ§θ⁰⊞υ⟦⟧ＵＭθＥι⊞Ｏ§υ⁺κμλ≦⮌υＩＥ§θ⁰Ｅθ⊟§υ⁺κμ

Try it online! Link is to verbose version of code. Outputs in Charcoal's default one-element-per-line format. Explanation:

Ｆ⁺θ§θ⁰⊞υ⟦⟧

Create a list of empty lists that will eventually hold the antidiagonals.

ＵＭθＥι⊞Ｏ§υ⁺κμλ

Extract the antidiagonals from the input. (This replaces each element in the array with the antidiagonal to which it belongs, but unfortunately I can't make use of this fact.)

≦⮌υ

Reverse the antidiagonals so that we can extract the elements in the original order using Pop().

ＩＥ§θ⁰Ｅθ⊟§υ⁺κμ`

Transpose the array, but read the elements back from the antidiagonals. (If Charcoal had an equivalent of JavaScript's unshift() then I would be able to unshift directly from the transposed array's antidiagonals to recover the elements.)

Answer 10

Haskell, 100 bytes

-3 bytes thanks to Wheat Wizard.

t m=[[m!!(i-1-x+y)!!(j+x-y)|j<-[0..l m-1],let[x,y]=min(i+j)<$>[l$m!!0,l m]]|i<-[1..l$m!!0]]
l=length

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An ugly port of my PARI/GP answer.

---

Curry (PAKCS), 101 bytes

-4 bytes thanks to Wheat Wizard.

t m=[[m!!(i-1-x+y)!!(j+x-y)|j<-[0..l m-1],let[x,y]=map(min$i+j)[l$m!!0,l m]]|i<-[1..l$m!!0]]
l=length

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Answer 11

Python3, 234 bytes:

E=enumerate
s=lambda m,d:(k for i,j in E(m)for x,k in E(j)if(i==d and x<(len(m[0])-d))or(i>d and x==(len(m[0])-1-d)))
r=lambda d,m:[[next(d[min(x,len(m)-1-y)])for y,_ in E(m)]for x,_ in E(m[0])]
lambda m:r({i:s(m,i)for i,_ in E(m)},m)

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