Problem 67. Find common elements in matrix rows

Sould be sort not flip, but I didn'r read the question properly ;)

Sould be sort not flip, but I didn'r read the question properly ;)

This is ingenious, man!

I think the Test Suite is broken for the results requiring an empty vector. Even though my function returns [ ] it is not detected as correct, as isequal() returns 0.

solving is easy, solving effectively is not! nice problem :)

This is far from elegant, but works! Is there a a vectorised way to approach this in a much more efficient way?

function y = common_by_row(x)
y=[];
[r,c]=size(x);
if r>0
a=x(1,:);
cnt=zeros(1,c);
for j=1:c
for i=1:r
if isempty(find(x(i,:)==x(1,j)))==0
cnt(j)=cnt(j)+1;
end
end
end
for j=1:c
if cnt(j)==r&&x(1,j)~=NaN&&ismember(x(1,j),y)==0
y=[y x(1,j)];
end
end
y=sort(y);
end
end

Why doesn't this work?? :( it returns an empty array as I specified...

Why is this incorrect? It's failing when the output is = [];
In matlab, I'm getting output as:
y =
1×0 empty double row vector
Is that not the same thing?

My code is correct but it is showing not a solution.......help me

The fact that an empty 0-by-1 matrix does not equal [] is a bit offputting.

I had to add an extra if-clause because the assertion isn't particularly forgiving about its definition of an "empty matrix".

The test suite should consider a 0x0 empty matrix (as given by y = []) as equivalent to a 1x0 empty matrix (which results from the code above it

solution uses recursion

nice !!!

argh! trying to debug third test case. I am returning a []. assert is comparing that to a []. How do I fail?
Both 3rd test case and 5th evaluate to [] locally, but I fail the cody asserts.

As others have later commented, the assert requires a 0×0 size 'empty' vector. See edited version in Solution 1276044.