# Python Little Goal 4

Python Day 10

1. #### Leap Year

I have a friend Cindy, she was actually born on Feb.29th, and I remember we celebrated her birthday together 2012 in the Maths place!
Leap Year was adapted in Gregorian Calendar.  There are Three Criteria:

• The year can be evenly divided by 4, is a leap year, unless:
• The year can be evenly divided by 100, it is NOT a leap year, unless:
• The year is also evenly divisible by 400. Then it is a leap year. pixel2013 / Pixabay

Solution:

```def is_leap(year):
if year % 4 != 0:
print("False")
elif year % 4 ==0 and year % 100==0 and year % 40==0:
print("Leap Year")
elif year % 4==0 and year % 100==0 and year % 40!=0 :
print("False")
else:
print("Leap Year")
```

Alternative Solutions:

```#Alternative Solution
def is_leap1(year):
return year % 4 == 0 and (year % 400 == 0 or year % 100 != 0)```

Output:

```print(is_leap(1900))
False```

#### 2. Second Largest Number in a List

Suppose we want to find the Second Largest Number in a List
Ex: List=[4,6,3,2,2] Output=4

Solution:

```arr=[2,5,4,2]
length=len(arr)
arr.sort()
print(arr[length-2])```

#### 3. Read an integer N, try to print out 123…NEX: N=3, Output 123

Solution:

```n= int(input())
print(*range(1,n+1), sep="")

n=8
1234567```

#### Summary:

we have been working with the Python commands Range, Length, Sort, Module (%), and  If-Then(elif-then, else), dfdAs all the books suggested, the Best way to learn to use the knowledge mindfully, so I organized this note 🙂

#### Happy Practice! 🦁

Reference:
https://www.hackerrank.com/

https://www.geeksforgeeks.org/python-largest-smallest-second-largest-second-smallest-list/

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