Given a string such as
s = '011110010000000100010111'
find the length of the longest string of consecutive 1's. In this example, the answer would be 4.
Input x = '110100111'
Output y is 3
Nice and useful !
Nice one, didn't think it was possible to do in one line initially...
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);
Solved on iPhone
Any suggestions on how to improve more ?
How can i improve the code?
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
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?
Return the 3n+1 sequence for n
Find the nearest integer
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