<prompt> Fumana mekhoa ea ho ntlafatsa snippet ea khoutu e latelang.def leap_yearselemo: haeba selemo 4 0 le selemo 100 ! 0 kapa selemo 400 0: kgutla Nnete e nngwe: kgutlisa Bohata  <response>: Instruction in English: Find ways to improve the following code snippet. . Response in English: 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_yearyear: 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 . Response in Sesotho: Tsela e ngoe ea ho ntlafatsa khoutu e tla ba ho e etsa e khutsoane haholoanyane ka ho khutlisetsa sephetho sa polelo e nang le maemo ka ho toba ntle le ho sebelisa sebopeho sa haeba se seng. Hape, tsebetso e ka rua molemo ka ho eketsoa ha docstring e hlalosang morero oa eona le tlhahiso e lebelletsoeng. Ena ke mofuta o ntlafalitsoeng oa khoutu: def leap_yearyear: Etsa qeto ea hore na selemo se fanoeng ke selemo se tlokomang ho latela almanaka ea Gregorian. :param year: Palo e emelang selemo. :return: Boolean e bontšang hore na selemo se fanoeng ke selemo se tlokomang. khutla selemo 4 0 le selemo 100 ! 0 kapa selemo 400 0 