I think it doesn't make it n^2. It is still linear - 2n, if you want. But constants are usually discarded in Big O, so just n. n^2 would be if, for example, for every element of that array there were another array which would be iterated.
STDERR:
error[E0433]: failed to resolve. Use of undeclared type or module `HashSet`
--> src/lib.rs:18:19
|
18 | let mut set = HashSet::new();
| ^^^^^^^ Use of undeclared type or module `HashSet`
error[E0433]: failed to resolve. Use of undeclared type or module `HashSet`
--> src/lib.rs:18:19
|
18 | let mut set = HashSet::new();
| ^^^^^^^ Use of undeclared type or module `HashSet`
error: aborting due to previous error
warning: build failed, waiting for other jobs to finish...
error: aborting due to previous error
error: Could not compile `qualified-rust`.
NOTE: Line numbers in error messages can be incorrect due to concatenation.
Nice code
I guess that Rust and Ts cases are different issues.
Same with me
'Map' only refers to a type, but is being used as a value here.
However on a local typescript / jasmine setup all tst cases pass.
O(n^2)
only happens when you have nested loops. Since the two loops are not nested at all (they're sequential), it'sO(n)
;-)I think it doesn't make it
n^2
. It is still linear -2n
, if you want. But constants are usually discarded in Big O, so justn
.n^2
would be if, for example, for every element of that array there were another array which would be iterated.Same error, did you report it? or is it our fault?
On playground works well.
Yes it is codewar sandbox problems.
Thanks for the explanation.
Please try my solution and tests on Rust playground:
Permalink to the playground
Direct link to the gist
Is there someone who also face on with the error?
It's weird. It's included at the top of the solution
use std::collections::HashSet;
Done some changes but the code was increased.
There is an error.
Nice, but this works as normal when the methods are called in the wrong order -- when it actually shouldn't.
Can't see errors. Could you elaborate?
Thanks for the answer.
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