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...
good problem.should think to solve
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);
no regexp is needed actually
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?
Project Euler: Problem 2, Sum of even Fibonacci
Reverse the elements of an array
Symmetry of vector
Number of Even Elements in Fibonacci Sequence
Determine if input is odd
Sums with Excluded Digits
Find the two-word state names
Find the numeric mean of the prime numbers in a matrix.
Remove any row in which a NaN appears
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