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I thought they were going to turn into Saddam Husseins.
Zero people in this post get the YanDev reference
if (!(number & 1))return trueis correct around half of the time
assert IsEven(2) == True assert IsEven(4) == True assert IsEven(6) == TrueAll checks pass. LGTM
import re def is_even(i: int) -> bool: return re.match(r"-?\d*[02468]$", str(i)) is not NoneCursed
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
trueWhat 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
You can just bitwise AND those with …000000001 (for however many bits are in your number). If the result is 0, then the number is even, and if it’s 1, then the number is odd. This works for negative numbers because it discards the negative signing bit.
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.
just check the least significant bit smh my head
If number%2 == 0: return("Even") Else: return("odd")When you sacrifice memory for an O(1) algorithm.
In this case still O(n)
Smh this is literally what switch statements are for
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
