1 should be an acceptable answer to n=3 as there a single way to triangulate a triangle - do not do anything. I don't get why you chose not to consider this.
Python.
I think it's important to say that n is the number of vertices the polygon has.
If the above is true, then the phrasing 'For n=2, possible_triangulations(2) returns 2,' Doesn't make any sense. To return 2, n should be 4; and when n = 2, it should return 'Not a valid polygon'.
Number of random tests is increased to 1001 to ensure all cases get tested. And your other issue also is fixed https://www.codewars.com/kumite/6830931d9b22cae2fe115973?sel=6830931d9b22cae2fe115973
Typo fixed in this fork: https://www.codewars.com/kumite/6830931d9b22cae2fe115973?sel=6830931d9b22cae2fe115973
Fixed in this fork (does not invalidate old solutions): https://www.codewars.com/kumite/6830931d9b22cae2fe115973?sel=6830931d9b22cae2fe115973
1 should be an acceptable answer to n=3 as there a single way to triangulate a triangle - do not do anything. I don't get why you chose not to consider this.
i am stupid ngl
Never thought to use min/max...and I always forget they exist LOL
Python.
I think it's important to say that n is the number of vertices the polygon has.
If the above is true, then the phrasing 'For n=2, possible_triangulations(2) returns 2,' Doesn't make any sense. To return 2, n should be 4; and when n = 2, it should return 'Not a valid polygon'.
should u check for int equality ? aq to question ? as i land with same solution
You can click View Solution under their post.
nice, i got a long way to go lol
Kata retired.
suboptimal solution.
Suggest your version of the correct code
Hermoso <3
same l0l0l0l
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