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Collections are a way for you to organize kata so that you can create your own training routines. Every collection you create is public and automatically sharable with other warriors. After you have added a few kata to a collection you and others can train on the kata contained within the collection.
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It's about expectations.
To cast a u32 to a char, you have to use
char::from_u32()
. This returns an option, making it explicit that the result may not be a valid unicode value, but when it does (result isSome(char)
, then you know it is what you expected it to be based on the value held by theu32
. The types match. This simply isn't the case with an i32, where you have a 50% chance to get it catastrophically wrong. So, as a programmer, you aren't allowed to simply cast signed to char that way. But it's your choice to "make a detour viau32
or smaller", because then it should be clear to you that casting a signed value to an unsigned value will yield "weird" results.Note that when casting to
u8
first, there is no possibility that you'll get an invalid unicode value because the entirety of 0-255 is valid, hence you're allowed to directly castu8 as char
Yeah, i see your point. But could one not argue that if
u32 as char
could panic depending on the u32 value, why is it any different than knowing the risk of a i32 panicking in the case of a -ve value for example?char
may not be what you think it is; it does not make sense to allow casting from a signed integer to a unicode scalar value. However, a signed integer can be cast to an unsigned of greater or equal size by simply interpreting the bits accordingly. Any unsigned up to 32bits can then be cast technically into achar
, however it might not map into the valid unicode range and panic, but at least as far as the type system is concerned it's possible to do.Had no idea you could chain casts like that – very nice! (though it seems redundant to not allow
i32 as char
?)This is a way to arounding up to nearest integer. For example 3/2 mast be 2, not 1.
"The restriction is that the characters in part1 and part2 should be in the same order as in s."
It should return False. The tests must not be very thorough!
Is it correct that this solution
isMerge "abc" "ca" "b"
returns True?You are not decreasing r by 5 but 4 that's the part I am having diffuculty to understand.
Can someone please explain in detail with an example, I couldn't understand the solution showed in the comments.
Found my problem, for some reason I mentally removed the parentheses (pls don't ask why), so my calculations are off by 1 (each iteration). I did
n - names.length - 1
, instead it wasn - (names.length - 1)
orn - names.length + 1
if you want.Sorry yeah, you're right. Fixing that the 21st should anyway endup on L(eonard) though:
So it doesn't change the fact that the index would go to
-1
.Your third line is wrong, it should be SSSS and so on
This comment is hidden because it contains spoiler information about the solution
I knew, that there will be very elegant solution, and as always it exceeded any of my expectations.
Thank you, it is beautiful.
Benefited greatly
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