Solution 1817575

Submitted on 16 May 2019 by Qinghua Ye

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Qinghua Ye on 16 May 2019

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

Sanzhar Askaruly on 17 Jun 2019

no regexp is needed actually

Adnan Anjum on 29 Jan 2020

Cool

Test	Status	Code Input and Output
1	Pass	x = '0'; y_correct = 0; assert(isequal(lengthOnes(x),y_correct))
2	Pass	x = '1'; y_correct = 1; assert(isequal(lengthOnes(x),y_correct))
3	Pass	x = '01'; y_correct = 1; assert(isequal(lengthOnes(x),y_correct))
4	Pass	x = '10'; y_correct = 1; assert(isequal(lengthOnes(x),y_correct))
5	Pass	x = '00'; y_correct = 0; assert(isequal(lengthOnes(x),y_correct))
6	Pass	x = '11'; y_correct = 2; assert(isequal(lengthOnes(x),y_correct))
7	Pass	x = '1111111111'; y_correct = 10; assert(isequal(lengthOnes(x),y_correct))
8	Pass	x = '100101011111010011111'; y_correct = 5; assert(isequal(lengthOnes(x),y_correct))
9	Pass	x = '01010101010101010101010101'; y_correct = 1; assert(isequal(lengthOnes(x),y_correct))
10	Pass	x = '0101010111000101110001011100010100001110110100000000110001001000001110001000111010101001101100001111'; y_correct = 4; assert(isequal(lengthOnes(x),y_correct))

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