115
Rule
(lemmy.blahaj.zone)
return true
is correct around half of the time
assert IsEven(2) == True
assert IsEven(4) == True
assert IsEven(6) == True
All checks pass. LGTM
just check the least significant bit smh my head
Just divide the number into its prime factors and then check if one of them is 2.
or divide the number by two and if the remainder is greater than
-(4^34)
but less than
70 - (((23*3*4)/2)/2)
then
true
What if the remainder is greater than the first, but not less than the latter?
Like, for example, 1?
Then you should return false, unless the remainder is also greater than or equal to the twenty second root of 4194304. Note, that I've only checked up to 4194304 to make sure this works, so if you need bigger numbers, you'll have to validate on your own.
i hate to bring this up, but we also need a separate function for negative numbers
I remember coding actionscript in Flash and using modulo (%) to determine if a number was even or odd. It returns the remainder of the number divided by 2 and if it equals anything other than 0 then the number is odd.
Yeah. The joke is that this is the obvious solution always used in practise, but the programmer is that bad that they don't know it and use some ridiculous alternative solutions instead.
I believe that's the proper way to do it.
import re
def is_even(i: int) -> bool:
return re.match(r"-?\d*[02468]$", str(i)) is not None
Cursed
If number%2 == 0: return("Even")
Else: return("odd")
When you sacrifice memory for an O(1) algorithm.
In this case still O(n)
Ask AI:
public static boolean isEven(int number) {
// Handle negative numbers
if (number < 0) {
number = -number; // Convert to positive
}
// Subtract 2 until we reach 0 or 1
while (number > 1) {
number -= 2;
}
// If we reach 0, it's even; if we reach 1, it's odd
return number == 0;
}Anything but using modulo I guess
This makes me happy that I don’t use genai
this post was submitted on 31 Oct 2024
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