Please tell the error in code below, as seems as perfectly fine as it's alternative.

Please tell why the below code is considered incorrect?

def is_leap(year):
    leap = False
    
    # Write your logic here
    if year%400==0 :
        leap = True
    elif year%4 == 0 and year%100 != 0:
        leap = True
    return leap

year = int(input())
print(is_leap(year))

If asked for, can provide link of where it has been marked as incorrect.

The alternative provided is :

def is_leap(year):
  leap = False

  if (year % 4 == 0) and (year % 100 != 0): 
      # Note that in your code the condition will be true if it is not..
      leap = True
  elif (year % 100 == 0) and (year % 400 != 0):
      leap = False
  elif (year % 400 == 0):
      # For some reason here you had False twice
      leap = True
  else:
      leap = False

  return leap

But, I see nothing gained, except redundant code.

Also, ran the stated (earlier) code, and gives correct results.

Please provide why the first code is incorrect?

First all, this is python.

Both of those work.

python
Python 3.11.8 (main, Feb 12 2024, 14:50:05) [GCC 13.2.1 20230801] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> def is_leap(year):
...     leap = False
...     
...     # Write your logic here
...     if year%400==0 :
...         leap = True
...     elif year%4 == 0 and year%100 != 0:
...         leap = True
...     return leap
... 
>>> year = int(input())
2010
>>> print(is_leap(year))
False

Second example has mixed indents.(Which wont work in python)

>>> def is_leap(year):
...     leap = False
...     
...     if (year % 4 == 0) and (year % 100 != 0): 
...         # Note that in your code the condition will be true if it is not..
...         leap = True
...     elif (year % 100 == 0) and (year % 400 != 0):
...         leap = False
...     elif (year % 400 == 0):
...         # For some reason here you had False twice
...         leap = True
...     else:
...         leap = False
...         
...     return leap
... 
>>> year = int(input())
2010
>>> print(is_leap(year))
False

The Python code for determining a leap year is "calendar.isleap(year)"

Anything else is redundant code.


If this is homework, then a correct answer should start by pointing out the built-in function, then solving it anyway, along with with a full set of test cases showing how the function handles all edge cases correctly.

The "alternative provided" is stupidly over-verbose but also looks like it is a correction of different code - so maybe the teacher sent you an updated version of someone else's attempt. Have you tried asking the teacher?

Originally Posted by grail Please provide why the first code is incorrect?

I don't know, the source says. If you allow, can quote the source.

Originally Posted by boughtonp The Python code for determining a leap year is "calendar.isleap(year)" Anything else is redundant code. If this is homework, then a correct answer should start by pointing out the built-in function, then solving it anyway, along with with a full set of test cases showing how the function handles all edge cases correctly. The "alternative provided" is stupidly over-verbose but also looks like it is a correction of different code - so maybe the teacher sent you an updated version of someone else's attempt. Have you tried asking the teacher?

The python code is:

def isleap(year):
    """Return True for leap years, False for non-leap years."""
    return year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)

I would suggest that if you are doing a course then it is expecting specific answers.
Whilst your version may work and still be correct, it would appear the source material required the answer provided

I actually prefer your version, as long as it can be proved to be correct using an appropriate set of test cases. I have doubts about the second with regard to the "% 400" case, since "% 100" is tested first, and it would also "pass" the "400 case." Very importantly, yours does not do that.

However, I would have written the final bit as: "else return false." I would not bother to have a "leap" variable – just "return true/false" – because you never subsequently test that variable.

The resulting code is clear: the most-restrictive case first, each one directly "returns" true or false, and the final "else" exists to very-clearly show the catch-all case.

But, such a function then needs to be rigorously tested against a set of appropriately-chosen dates which test both the expected results and the "edge cases."