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def even(n: int) -> bool: code = "" for i in range(0, n+1, 2): code += f"if {n} == {i}:\n out = True\n" j = i+1 code += f"if {n} == {j}:\n out = False\n" local_vars = {} exec(code, {}, local_vars) return local_vars["out"]scalable version
No, no, you should group the
return falselines together 😤😤if (number == 1) return false; else if (number == 3) return false; else if (number == 5) return false; //... else if (number == 2) return true; else if (number == 4) return true; //...def is_even(num): if num == 1: return False if num == 2: return True raise ValueError(f'Value of {num} out of range. Literally impossible to tell if it is even.')I’m partial to a recursive solution. Lol
def is_even(number): if number < 0 or (number%1) > 0: raise ValueError("This impl requires positive integers only") if number < 2: return number return is_even(number - 2)I prefer good ole regex test of a binary num
function isEven(number){ binary=$(echo "obase=2; $number" | bc) if [ "${binary:-1}" = "1" ]; then return 255 fi return 0 }
