Isn't it annoying when you're taking a picture, but the background detracts from the actual substance of the image? I'd say it is. I need to know how much I should crop so that I get rid of this problem! But - as usual - I am quite lazy, so I need someone to do this for me...

Task & Rules

Given a binary matrix representing the image, output the dimensions (width and height) of the smallest sub-matrix that contains all the $1$s in the original matrix. A sub-matrix is a block of adjacent entries from the original matrix. Equivalently, it is a new matrix formed by overlapping a subset of adjacent rows and a subset of adjacent columns of the original.

• It is allowed to take the width and the height of the matrix as input as well.


• The input is guaranteed to contain at least one $1$.


• You can take input and provide output through any standard method, while taking note that these loopholes are forbidden by default. This is code-golf, so try to complete the task in the least bytes you can manage in your language of choice.

Example

$$left[beginmatrix
colorred0&colorred0&colorred0&colorred0&colorred0&colorred0\
colorred0&colorblue1&colorblue0&colorblue1&colorblue0&colorblue0\
colorred0&colorblue1&colorblue1&colorblue0&colorblue1&colorblue1\
colorred0&colorblue0&colorblue1&colorblue0&colorblue1&colorblue0\
colorred0&colorred0&colorred0&colorred0&colorred0&colorred0endmatrixright] longrightarrow left[beginmatrix1&0&1&0&0\1&1&0&1&1\0&1&0&1&0endmatrixright]longrightarrow(5,3)$$

Test cases

Input | Output

[[0,1,0,0,0,1,0]]
--> (5,1) or (1,5)

[[0,0,0,0,0],[0,1,0,1,0],[0,0,1,0,0]]
--> (3,2) or (2,3)

[[1,1,1,1],[0,0,0,0],[0,0,0,0],[1,0,0,0]]
--> (4,4)

[[0,0,0,0,0,0],[0,1,0,1,0,1],[0,0,0,0,0,0]]
--> (5,1) or (1,5)

[[0,0,0,0,0],[0,1,0,1,0],[0,0,1,0,0],[0,1,0,1,0],[0,0,0,0,0]]
--> (3,3)

[[0,0,0,0,0,0],[0,1,0,1,0,0],[0,1,1,0,1,1],[0,0,1,0,1,0],[0,0,0,0,0,0]]
--> (5,3) or (3,5)

Octave, 57 56 bytes

Here find does the heavy lifting: It finds the row and colum indices of the nonzero entries. Then we just have to find the difference between the maximum and the minimum (plus one) for each of those seperately.

Thanks @beaker and @AndrasDeak for -1 byte!

@(a)[I1:2]=find(a),cellfun(@(x)max(x)-min(x)+1,I)2

Try it online!

Retina 0.8.2, 83 bytes

+`^0+¶|¶0+$

+1`((.)*).(¶(?<-2>.)*).(?<=(1)¶.*|(.))
$1$3$4$5
(¶?)*0*(.*1)0*
$#1 $.2

Try it online! Explanation:

+`^0+¶|¶0+$

Delete leading and trailing zero rows.

+1`((.)*).(¶(?<-2>.)*).(?<=(1)¶.*|(.))
$1$3$4$5

Remove all 0s on the lines above the last. Remove all the 1s two, but change the digit underneath on the next line to a 1 in that case. This bitwise or's the rows together.

(¶?)*0*(.*1)0*
$#1 $.2

Count the number of rows as the number of newlines plus 1 and the number of columns as the number of digits between the first and last 1.

Japt, 16 bytes

[UUy]
®=ðd)ÌaZÎÉ

Try it

Explanation

[ ] :Create an array containing
 U : The input and
 Uy : The input transposed
n :Reassign to U
® :Map each Z
 ð : Indices of elements where
 d : Any element is truthy (not zero)
 = ) : Reassign to Z
 Ì : Last element
 a : Absolute difference with
 ZÎ : First element
 É : Minus 1