Pfthb

def leap_years(start, end):
 if start < 1500 or start > 2100:
 return
 if end < 1500 or end > 2100:
 return
 i = 0
 for i in range(start, end + 1):
 if i % 4 == 0 and (i % 100 != 0 or i % 400 == 0):
 print(i)


print(leap_years(1998, 2008))

I want to output be 3 and not 2000,2004,2008

This is a good place to consider a list comprehension to create a list of leap years:

def leap_years(start, end):
 if start < 1500 or start > 2100:
 return
 if end < 1500 or end > 2100:
 return

 leaps = [ i for i in range(start, end+1) if i % 4 == 0 and (i % 100 != 0 or i % 400 == 0) ]

 print( len(leaps) )

leap_years(1998,2008)

Result is:
3

You need to add a counter:

def leap_years(start, end):
 if start < 1500 or start > 2100:
 return 0
 if end < 1500 or end > 2100:
 return 0
 i, count = 0, 0
 for i in range(start, end + 1):
 if i % 4 == 0 and (i % 100 != 0 or i % 400 == 0):
 count += 1
 return count

print(leap_years(1998, 2008))

Output

3

In the above code the counter is the variable count each time the leap year condition is met it increments by 1. Note that now the function leap_years returns the amount of leap years.

We can use sum(...) here to count the number of elements:

def leap_years(start, end):
 if not 1500 < start < 2100:
 return 0
 if not 1500 < end < 2100:
 return 0
 return sum(
 (not y % 4 and y % 100 != 0) or not y % 400
 for y in range(start, end + 1)
 )

This works since in Python True is equal to 1, and False to 0, so we count the number of elements.

The above is however not very efficient: we can in fact calculate the number of leap years over huge ranges, without having to iterate over them.

We can for example calculate the number of elements dividably by 4 between a and b (both inclusive), by calculating ((b - c)/4) + 1 where c is the next element dividable by 4.

With the same logic we can exclude the number of elements dividable by 100, and include the ones dividably by 400, like:

def num_div(div, frm, to):
 return max(0, (to - (frm + (div - frm) % div) + div) // div)

def leap_years(start, end):
 return num_div(4, start, end) - num_div(100, start, end) + num_div(400, start, end)

For example, given the Gregorian calendar with these rules always existed, the number of leap years between 1234 AD and 123'456'789 AD is:

>>> leap_years(1234, 123456789)
29937972