If p is the perimeter of a right angle triangle with integral length sides, { a, b, c }, there are exactly three solutions for p = 120.
{[20,48,52], [24,45,51], [30,40,50]}
Given any value of p, return a cell array whose elements are the sorted side lengths of a possible right triangle whose perimeter is p. Furthermore, the elements of the output should be sorted by their shortest side length.
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Hi Team,
I am getting error after submitting as,
"We're sorry, but something went wrong. We've been notified about this issue and we'll take a look at it shortly"
But completion bar has been incremented.
Thanks and Have a Good DaY
Well I don't know why I can't pass the test
could anyone help
function c = right_triangle_sides(p)
c = {};
k=1;
for i=1:ceil(p/3)
for j=i:ceil((p-i)/2)
if i^2+j^2==(p-i-j)^2
c(k)=mat2cell([i j p-i-i],1);
k=k+1;
end
end
end
end
To creator of this group,
Please check the test case 3 in the Problem of calculating side length from area of triangle. One ';' is missing which is not allowing the case to pass for correct solution.
This problem really highlights problems with the Cody scoring system. The score using a double for loop to brute force the lengths beats out a single for loop using equations. I tested my code in Matlab vs double loops and it's faster than most double loops I tested by several orders of magnitude for larger numbers (p>1000). There is also some sort of caching going on where my code gets orders of magnitude faster again if ran a few times vs most double for loops.
There is one exception with a double loop that is really fast by Hung Hoang. They use the for loops in a way that doesn't make it O^2.
interesting problem
đánh giá là bài lỗi vãi đái:))