function y = lengthOnes(x)
a = [];
idx = [1,find(x=='0') + 1,length(x)+2];
le = length(idx);
a(1:le-1) = idx(2:le)-idx(1:le-1)-1;
y = max(a);
end

no regexp is needed actually

Cool

Solved on iPhone

Any suggestions on how to improve more ?

How can i improve the code?

a=regexp(x,'0+','split')
a=a{max(length(a))}
y=length(a)
With this code, every test passes except 4th one. Why is it so?

any way of improving this algorithm?
when y is empty vector, max(y) is not returning zero which is making my code complicated

It runs well in my Matlab program.

How much time it takes in your MATLAB? There is a limit of 50 seconds for Cody

probably the most inefficient code ever written :)

I'm loving it

efficient idea

I know it's not very efficient, but I sort of liked an idea behind this loop. Hence shared.

Great solution that's easy to understand adn translatable to future use.

extremely unfamiliar with "regexp" function results in a extremely stupid code...

hmmmmm.... ;-) nice

dynamic expression! this is super cool, i didnt know matlab had this and that you could write a matlab code that's so hard to read. Why the 49?

Honestly, right now I find it hard to read as well. 49 is the decimal representation of the ASCII character '1' (type +'1' in the Matlab Command Window).

finally a proper use of regexp in Cody

can you explain your code?