I thought it would be really cool if you could save the amount of time a solution took in your database and use this information to rank the solutions.
Since the one that took the least amount of time is probably the most efficient one.
anOver(n) with parameter n: returns an array (n terms) of the a(n)/n for every n such g(n) != 1 (not tested but interesting result)
The n represents two different things which can be confusing.
The n in anOver(n) represents the size of the array
but the n in a(n)/n represents the index.
Maybe you could do something like
anOver(n) with parameter n: returns an array (n terms) of the a(i)/i for every i such g(i) != 1 (not tested but interesting result)
Clever
Clever, but modulo operator can return negative values, so this isn't always correct.
Not anymore ;-)
It's already there
Good advice, ZozoFouchtra. Thanx!
Is there anyway to retranslate the java...version
Could someone explain to me where did the 5 come from?
well its for beginners so maybe its intentional.
True but this one has more complexity so maybe rename it to valid-braces level 2 or something
or maybe make a reference in the description.
Modified. Thanks!
I thought it would be really cool if you could save the amount of time a solution took in your database and use this information to rank the solutions.
Since the one that took the least amount of time is probably the most efficient one.
anOver(n) with parameter n: returns an array (n terms) of the a(n)/n for every n such g(n) != 1 (not tested but interesting result)
The n represents two different things which can be confusing.
The n in anOver(n) represents the size of the array
but the n in a(n)/n represents the index.
Maybe you could do something like
anOver(n) with parameter n: returns an array (n terms) of the a(i)/i for every i such g(i) != 1 (not tested but interesting result)
no test
no test case
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