First, we will solve this problem when ~ N ~ is a positive integer.

In this case, simply print ~ N - 1 ~, since the condition is strictly less than.

Next, solve the problem if ~ N \le 0 ~.

Now, if ~ N ~ is not positive, there are 0 positive integers less than ~ N ~. In this case, use a simple conditional or a max function to output ~ \text{max}(0, N - 1) ~.

Finally, if ~ N ~ is not an integer, then the integer part of ~ N ~ is included in the list, so we should output the integer part of ~ N ~. However, if ~ N ~ is an integer, we should still output ~ N - 1 ~. One way to solve this is to use a ceil function, so that you get the first number greater than or equal to ~ N ~ which is an integer.

Therefore, in total, you should output ~ \text{max}(\lceil N \rceil - 1, 0) ~.