196

import re

def is_even(i: int) -> bool:
    return re.match(r"-?\d*[02468]$", str(i)) is not None

If number%2 == 0: return("Even")
Else: return("odd") 

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;
}

When you sacrifice memory for an O(1) algorithm.

In this case still O(n)

Just divide the number into its prime factors and then check if one of them is 2.