return trueis correct around half of the time
assert IsEven(2) == True assert IsEven(4) == True assert IsEven(6) == TrueAll 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.
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 NoneCursed
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
