CODEWITHSUNDEEP
python-3.x
shiven_guru
shiven_guru

Reputation: 1

how To print the list of next 15 leap years from a given year?

Write a Python program to generate the next 15 leap years starting from a given year. Populate the leap years into a list and display the list. Also write the pytest test cases to test the program.

def find_leap_years(given_year):
    list_of_leap_years=[0]*15

    # Write your logic here
    if given_year%100==0 and given_year%400!=0:
        for i in range(0,15):
             temp=given_year+(4*(i+1))
             list_of_leap_years[i]=temp
    elif given_year%400==0:
        for i in range(0,15):
            temp=given_year+(4*i)
            list_of_leap_years[i]=temp
    elif given_year%4==0:
        for i in range(0,15):
            temp=given_year+(4*i)
            list_of_leap_years[i]=temp
    elif given_year%4==1:
        for i in range(0,15):
            temp=given_year+3+(4*i)
            list_of_leap_years[i]=temp
    elif given_year%4==2:
        for i in range(0,15):
            temp=given_year+2+(4*i)
            list_of_leap_years[i]=temp
    elif given_year%4==3:
        for i in range(0,15):
            temp=given_year+1+(4*i)
            list_of_leap_years[i]=temp

    return list_of_leap_years

next_leap_years=find_leap_years(1684)
print(next_leap_years)

when I take given year as 1684, the test case is failing, as my program prints 1700 in leap years list, but 1700 isn't a leap year.

Upvotes: 0

Views: 7971

Answers (4)

user19563477
user19563477

Reputation: 1

  def find_leap_years(given_year):
     count = 0
     leap_years = []


     while count < 15:
         if given_year % 400 == 0  or (given_year % 4 == 0 and given_year % 100 != 0):
                leap_years.append(given_year)
                count += 1
                given_year +=4
         else:
           given_year +=1
     return leap_years
leap_years_list = find_leap_years(2000)
print(leap_years_list)

Upvotes: 0

SATYAM TRIPATHI
SATYAM TRIPATHI

Reputation: 67

import calendar

def Leap_Year(Test_input):
    count = 0
    while count < 15:
        if calendar.isleap(Test_input):
            print(Test_input)
            count += 1
        Test_input += 1


Test_input = int(input("Enter Year: "))
Leap_Year(Test_input)

Upvotes: 1

Pranav Void
Pranav Void

Reputation: 357

def find_leaps(year, counts=15):
     cnt = 0
     leaps = []
     while cnt != counts:
         if is_leap(year):
             leaps.append(year)
             cnt += 1
         year += 1
     return leaps

and 

 def is_leap(year):
     return (year%4 == 0 and year%100 != 0) or (year % 400 == 0)

so 

In [3]: find_leaps(1684)
Out[3]:
[1684,
 1688,
 1692,
 1696,
 1704,
 1708,
 1712,
 1716,
 1720,
 1724,
 1728,
 1732,
 1736,
 1740,
 1744]

Upvotes: 2

Andrej Kesely
Andrej Kesely

Reputation: 195528

I would try something simpler and utilize datetime module. When constructing the date of 29th February fails, the exception is thrown - and we can catch it:

from datetime import date

def find_leap_years(year, num=15):
    count = 0
    while count < num:
        try:
            d = date(year, 2, 29)
            yield year
            count += 1
        except ValueError:
            continue
        finally:
            year += 1

for i, years in enumerate( find_leap_years(1690, 15), 1 ):
    print(i, years)

Prints (note that 1700 isn't among the returned values):

1 1692
2 1696
3 1704
4 1708
5 1712
6 1716
7 1720
8 1724
9 1728
10 1732
11 1736
12 1740
13 1744
14 1748
15 1752

Upvotes: 2

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