def leap_year(year): 
    if (year % 4 == 0 and year % 100 != 0) or year % 400 == 0: 
        return True 
    else: 
        return False
Find ways to improve the following code snippet.
One way to improve the code would be to make it more concise by returning the result of the conditional statement directly without using an if-else structure. Also, the function could benefit from the addition of a docstring that explains its purpose and expected input. Here is an improved version of the code:

def leap_year(year):
    """
    Determine if a given year is a leap year according to the Gregorian calendar.

    :param year: An integer representing the year.
    :return: A boolean indicating whether the given year is a leap year.
    """
    return (year % 4 == 0 and year % 100 != 0) or year % 400 == 0