Remove surrounding zeroes of a 2d array

Question

This is a 2-dimensional version of this question.

Given a non-empty 2-dimensional array/matrix containing only non-negative integers:

$$ \begin{bmatrix} {\color{Red}0} & {\color{Red}0} & {\color{Red}0} & {\color{Red}0} & {\color{Red}0} \\ {\color{Red}0} & {\color{Red}0} & 0 & 1 & 0 \\ {\color{Red}0} & {\color{Red}0} & 0 & 0 & 1 \\ {\color{Red}0} & {\color{Red}0} & 1 & 1 & 1 \\ {\color{Red}0} & {\color{Red}0} & {\color{Red}0} & {\color{Red}0} & {\color{Red}0} \end{bmatrix} $$

Output the array with surrounding zeroes removed, i.e. the largest contiguous subarray without surrounding zeroes:

$$ \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \end{bmatrix} $$

Examples:

$$ \begin{bmatrix} {\color{Red}0} & {\color{Red}0} & {\color{Red}0} & {\color{Red}0} & {\color{Red}0} \\ {\color{Red}0} & {\color{Red}0} & 0 & 1 & 0 \\ {\color{Red}0} & {\color{Red}0} & 0 & 0 & 1 \\ {\color{Red}0} & {\color{Red}0} & 1 & 1 & 1 \\ {\color{Red}0} & {\color{Red}0} & {\color{Red}0} & {\color{Red}0} & {\color{Red}0} \end{bmatrix} \mapsto \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \end{bmatrix} $$

Input:
[[0, 0, 0, 0, 0], [0, 0, 0, 1, 0], [0, 0, 0, 0, 1], [0, 0, 1, 1, 1], [0, 0, 0, 0, 0]]

Output:
[[0, 1, 0], [0, 0, 1], [1, 1, 1]]

$$ \begin{bmatrix} {\color{Red}0} & {\color{Red}0} & {\color{Red}0} & {\color{Red}0} \\ {\color{Red}0} & 0 & 0 & 3 \\ {\color{Red}0} & 0 & 0 & 0 \\ {\color{Red}0} & 5 & 0 & 0 \\ {\color{Red}0} & {\color{Red}0} & {\color{Red}0} & {\color{Red}0} \end{bmatrix} \mapsto \begin{bmatrix} 0 & 0 & 3 \\ 0 & 0 & 0 \\ 5 & 0 & 0 \end{bmatrix} $$

Input:
[[0, 0, 0, 0], [0, 0, 0, 3], [0, 0, 0, 0], [0, 5, 0, 0], [0, 0, 0, 0]]

Output:
[[0, 0, 3], [0, 0, 0], [5, 0, 0]]

$$ \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \mapsto \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} $$

Input:
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]

Output:
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]

$$ \begin{bmatrix} {\color{Red}0} & {\color{Red}0} & {\color{Red}0} & {\color{Red}0} \\ {\color{Red}0} & {\color{Red}0} & {\color{Red}0} & {\color{Red}0} \\ {\color{Red}0} & {\color{Red}0} & {\color{Red}0} & {\color{Red}0} \end{bmatrix} \mapsto \begin{bmatrix} \end{bmatrix} $$

Input:
[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]

Output:
[]

$$ \begin{bmatrix} {\color{Red}0} & {\color{Red}0} & {\color{Red}0} & {\color{Red}0} \\ 1 & 1 & 1 & 1 \\ {\color{Red}0} & {\color{Red}0} & {\color{Red}0} & {\color{Red}0} \end{bmatrix} \mapsto \begin{bmatrix} 1 & 1 & 1 & 1 \end{bmatrix} $$

Input:
[[0, 0, 0, 0], [1, 1, 1, 1], [0, 0, 0, 0]]

Output:
[[1, 1, 1, 1]]

$$ \begin{bmatrix} {\color{Red}0} & 1 & {\color{Red}0} & {\color{Red}0} \\ {\color{Red}0} & 1 & {\color{Red}0} & {\color{Red}0} \\ {\color{Red}0} & 1 & {\color{Red}0} & {\color{Red}0} \end{bmatrix} \mapsto \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} $$

Input:
[[0, 1, 0, 0], [0, 1, 0, 0], [0, 1, 0, 0]]

Output:
[[1], [1], [1]]

$$ \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 1 \\ 1 & 1 & 1 & 1 \end{bmatrix} \mapsto \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 1 \\ 1 & 1 & 1 & 1 \end{bmatrix} $$

Input:
[[1, 1, 1, 1], [1, 2, 3, 1], [1, 1, 1, 1]]

Output:
[[1, 1, 1, 1], [1, 2, 3, 1], [1, 1, 1, 1]]

Answer 1

05AB1E (legacy), 13 bytes

2Fζ2FRDv¬O_i¦

Try it online or verify all test cases.

Explanation:

2F                    # Loop two times:
  ζ                   #  Zip/transpose; swapping rows/columns
                      #  (takes the input-matrix implicitly in the first iteration)
   2F                 #  Inner loop two times:
     R                #  Reverse the rows
      Dv              #  Inner loop over the rows:
        ¬             #   Get the first row (without popping the matrix)
         O_i          #   If the row consists only of 0s:
            ¦         #    Remove this first row from the matrix
                      # (output the result implicitly)

Answer 2

Attache, 40 bytes

Fixpoint[{Flip!_[N[0=Sum!_@0]...#_]}//4]

Try it online! Same business as below, just a bit smarter, splitting the process into four steps instead of two.

Alternatives

47 bytes: Fixpoint[{Reverse=>Tr!_[N[0=Sum!_@0]...#_]}//4]

48 bytes: Fixpoint[{MatrixRotate!_[N[0=Sum!_@0]...#_]}//4]

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Attache, 68 bytes

Fixpoint[{n.=Dim@_@-1Tr[{_@1!in~-n'0or Sum!_@0}\Enumerate@_<:0]}//2]

Try it online!

Twice: This removes any first or last row whose sum is 0, then transposes the array. Then, this process is repeated until the result does not change.

Answer 3

JavaScript (Node.js), 100 bytes

f=(a,i=4,b=a[0])=>b&&i?/[1-9]/.test(b)?f([...b].map((c,j)=>a.map(d=>d.pop())),i-1):f(a.slice(1),i):a

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