Removed J-brackets

Question

For a 2 dimensional array we will call the elements in either the first row or the last column the "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] \$

Your task is to take as input a 2 dimensional array of positive integers, and repeatedly remove J-brackets from it until the remaining array has no more elements. Your output should be all the J-brackets removed in this process in the order they were removed.

The input will always be perfectly rectangular, and both dimensions will be at least 1. You may take input in any reasonable format.

This is code-golf so the goal is to minimize the size of your source code as measured in bytes.

Test cases

[[2]] -> [[2]]
[[1,2],[3,4]] -> [[1,2,4],[3]]
[[1,2,4,8],[9,3,6,7],[3,3,2,9]] -> [[1,2,4,8,7,9],[9,3,6,2],[3,3]]
[[1,2],[3,4],[5,6]] -> [[1,2,4,6], [3,5]]
[[1,2],[3,4],[5,6],[7,9]] -> [[1,2,4,6,9], [3,5,7]]

Answer 1

Haskell, (53) 47 bytes

Fixed and golfed 6 bytes thanks to Wheat Wizard

f((a:b):c)=(a:b++(last<$>c)):f(init<$>c)
f x=[]

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Answer 2

R, 37 bytes

\(m)split(m,pmin(row(m),rev(col(m))))

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Working on the related challenge led me to this approach. Test harness taken from pajonk's answer.

In an \$m\times n\$ matrix \$A\$, each element \$A_{ij}\$ is in the \$p^\text{th}\$ J-bracket if and only if \$\min(i,n+1-j)=p\$.

R has some odd built-ins that return the matrix \$R=\text{row}(M)\$ where \$R_{ij}=i\$ and similarly for \$C=\text{col}(M)\$. Reversing the column matrix luckily performs the right operation, and we take the parallel minimum of these matrices to obtain a matrix of J-brackets, which split helpfully breaks into groups of the right order.

Answer 3

BQN, 11 bytes^SBCS

⌊⌜⟜⌽○↕´∘≢⊸⊔

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          ⊔  # Group the values by:
        ≢    # The shape of the matrix
      ´      # Reduce the shape by:
    ○↕       # Convert both integers to a range [0, n)
  ⟜⌽         # Reverse the right range
⌊⌜           # Minimum table

Answer 4

R, 117 88 86 76 72 bytes

Edit: -29 31 bytes thanks to @Giuseppe; -10 bytes thanks to Robin Ryder; another -4 bytes thanks to @Giuseppe.

f=\(M)"if"(1%in%dim(t(M)),M,list(c(M[1,],M[-1,n<-ncol(M)]),f(M[-1,-n])))

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Recursive approach with some clever tricks by @Giuseppe and Robin Ryder.

Answer 5

Python 2, 52 bytes

a=input()
while[[]]<a:print a.pop(0)+map(list.pop,a)

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-4 bytes thanks to @ovs

Answer 6

Jelly, 10 bytes

^{ßest Friends Forever?!}

Ḣ;Ṫ€W;ßFF?

A monadic Link that accepts a rectangular list of lists and yields a list of lists.

Try it online! Or see the test-suite.

How?

Ḣ;Ṫ€W;ßFF? - Link: list of lists, X
Ḣ          - head -> top row of X (mutates X)
   €       - for each of (the remaining rows of) X:
  Ṫ        -   tail -> last element of the row (mutates X)
 ;         - concatenate these together
    W      - wrap that list in a list -> Z
         ? - if...
        F  - ...condition: flatten (the mutated) X (falsey once empty or only empty rows)
      ß    - ...then: call this Link with (the mutated) X
       F   - ...else: flatten (the mutated) X -> an empty list
     ;     - Z concatenate that

Answer 7

K (ngn/k), 44 bytes

{,'/(&/d)#'(-1 1*!'d:#'1*:\x)_''1(1_'|+:)\x}

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Solved in my worst attempt at find-the-j-twin. It seems a shame to let it go to waste...

Answer 8

Vyxal, 14 bytes

λ[ḣƛt;J,ḢƛṪ;x¤

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Explanation

λ[ḣƛt;J,ḢƛṪ;x¤
λ                 Open a lambda (for recursion)
 [                If statement (check if truthy)
  ḣ               Separate the first item, push both sides to stack
   ƛt;            Map to last item of each
      J           Join both lists
       ,          Pop and print (the J bracket)
        Ḣ         All but the first item
         ƛṪ;      Map to all but the last item of each
            x     Recurse
             ¤    Empty space (so that it won't output an empty list at the end, since stack is implicitly output)

Answer 9

Python, 49 bytes

f=lambda a,i=0:a and[*a.pop(i),*f([*zip(*a)],~i)]

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Loosely based on @pxeger's Python 2 answer.

Answer 10

PARI/GP, 51 bytes

f(m)=m&&print(concat(m[1,],m[^1,#m]~))*f(m[^1,^#m])

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Answer 11

APL+WIN, 47 bytes

Prompts for 2 dimensional matrix

m←⎕⋄⍎∊(⌊/⍴m)⍴⊂'m[1;],1↓,m[;1↓⍴m]⋄m←1 0↓0 ¯1↓m⋄'

Try it online! Thanks to Dyalog APL Classic

Answer 12

05AB1E, 15 bytes

[ćs©€θ«,®€¨Wg_#

Try it online or verify all test cases.

Explanation:

[          # Loop indefinitely:
 ć         #  Extract head; pop and push first row and remainder-matrix
  s        #  Swap so the remainder-matrix is at the top
   ©       #  Store it in variable `®` (without popping)
    €θ     #  Pop and only leave the last value of each row
      «    #  Merge it to the extracted first row
       ,   #  Pop and output this list with trailing newline
   ®       #  Push matrix `®` again
    ۬     #  Remove the last integer from each row
      W    #  Push the flattened minimum of the matrix (without popping)
       g   #  Pop and push the length of it
        _  #  If this length is 0:
         # #   Stop the infinite list

Answer 13

Java 8, 109 bytes

m->{for(int i=0,j;;){var t=m[j=i++];for(;++j<m.length;)t.add(m[j].pop());if(t.size()>0)System.out.print(t);}}

Input as an array of Integer-Stacks; output prints directly to STDOUT.

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Answer 14

Charcoal, 20 bytes

≔⮌ＡθＷ›θＥθυＩ⟦⁺⊟θＥ⮌θ⊟κ

Try it online! Link is to verbose version of code. Explanation: Port of @pxeger's Python solution.

≔⮌Ａθ

Reverse the input as Charcoal can only pop from the end of a list.

Ｗ›θＥθυ

Repeat until the input has no rows or columns.

Ｉ⟦⁺⊟θＥ⮌θ⊟κ

Remove the now last row and the last column of the remaining rows reversed back into their original order and output those values.

Answer 15

Ruby, 46 bytes

f=->m{[m.shift+m.map(&:pop),*m*"">""?f[m]:[]]}

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Takes advantage of shift and pop which modifies object in place and return what we needed.

Checking the updated m was a bit difficult because pop and shift may return nil which is truthy in Ruby

Answer 16

Factor, 69 bytes

[ [ 1 cut flip 1 short cut* rot prepend concat . flip ] until-empty ]

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Explanation

It's a quotation (anonymous function) that takes a matrix as input and prints its J-brackets, one per line.

• [ ... ] until-empty Repeatedly call a quotation on an input until it is empty.

         ! { { 1 2 4 8 } { 9 3 6 7 } { 3 3 2 9 } }
1 cut    ! { { 1 2 4 8 } } { { 9 3 6 7 } { 3 3 2 9 } }
flip     ! { { 1 2 4 8 } } { { 9 3 } { 3 3 } { 6 2 } { 7 9 } }
1 short  ! { { 1 2 4 8 } } { { 9 3 } { 3 3 } { 6 2 } { 7 9 } } 1
cut*     ! { { 1 2 4 8 } } { { 9 3 } { 3 3 } { 6 2 } } { { 7 9 } }
rot      ! { { 9 3 } { 3 3 } { 6 2 } } { { 7 9 } } { { 1 2 4 8 } }
prepend  ! { { 9 3 } { 3 3 } { 6 2 } } { { 1 2 4 8 } { 7 9 } }
concat   ! { { 9 3 } { 3 3 } { 6 2 } } { 1 2 4 8 7 9 }
.        ! { { 9 3 } { 3 3 } { 6 2 } }
flip     ! { { 9 3 6 } { 3 3 2 } }

Answer 17

Wolfram Language (Mathematica), 91 bytes

(l=Length@#&@@#;Flatten@Pick[#,MapIndexed[Min[#,l-#2+1]&@@#2&,#,{2}],i]~Table~{i,Tr[1^#]})&

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Feels too long but I couldn't find anything shorter

Answer 18

JavaScript (ES6),  66 63  59 bytes

Saved 4 bytes thank to @l4m2 (on both versions)

Returns a list.

f=m=>[[...m.shift(),...m.map(r=>r.pop())],...m>'0'?f(m):[]]

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Commented

f =                // f is a recursive function taking
m =>               // a matrix m[]
[                  //
  [ ...m.shift(),  // extract the first row and split it
    ...m.map(r =>  // for each row r[] in m[]:
      r.pop()      //   extract the last element of r[]
    )              // end of map()
  ],               //
  ...m > '0' ?     // if there's still at least one number in m[]:
    f(m)           //   append the result of a recursive call
  :                // else:
    []             //   stop the recursion
]                  //

---

JavaScript (V8),  65 62  58 bytes

Prints the results.

f=m=>print([[m.shift()],...m].map(r=>r.pop()))|m>'0'&&f(m)

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Answer 19

PHP 5.4 (93 chars)

The function j take a $m PHP array and return the answer as a PHP array.

function j($m){for(;$k=array_shift($m);$j[]=$k)foreach($m as&$z)$k[]=array_pop($z);return$j;}

Variant using 105 characters :

The function j take a $m matrix and print the answer as JavaScript Object Notation array.

function j($m){for(;$k=array_shift($m);$j[]=$k)foreach($m as&$z)$k[]=array_pop($z);echo json_encode($j);}

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