# C++ Program to Check Leap Year

A year consisting of 366 days instead of the usual 365 days is a leap year. Every fourth year is a leap year in the Gregorian calendar system. In this article, we will learn how to write a C++ program to check leap year.

A year is a leap year if one of the following conditions is satisfied:

1. The year is a multiple of 400.
2. The year is a multiple of 4 but not a multiple of 100.

## Algorithm to Check Leap Year

The algorithm implements the conditions specified above to check for leap year.

```if (year % 400 = 0)
return true (Leap year)
else if (year % 100 = 0)
return false (Not a leap year)
else if (year % 4 = 0)
return true (Leap year)
else
return false (Not a leap year)
endif```

## C++

 `// C++ program to check if a given` `// year is a leap year or not` `#include ` `using` `namespace` `std;`   `// Function to check leap year` `bool` `checkYear(``int` `year)` `{` `    ``if` `(year % 400 == 0) {` `        ``return` `true``;` `    ``}`   `    ``// not a leap year if divisible by 100` `    ``// but not divisible by 400` `    ``else` `if` `(year % 100 == 0) {` `        ``return` `false``;` `    ``}`   `    ``// leap year if not divisible by 100` `    ``// but divisible by 4` `    ``else` `if` `(year % 4 == 0) {` `        ``return` `true``;` `    ``}`   `    ``// all other years are not leap years` `    ``else` `{` `        ``return` `false``;` `    ``}` `}`   `// Driver code` `int` `main()` `{` `    ``int` `year = 2000;`   `    ``checkYear(year) ? cout << ``"Leap Year"` `                    ``: cout << ``"Not a Leap Year"``;`   `    ``return` `0;` `}`

Output

`Leap Year`

### Complexity Analysis

• Time Complexity: Since there are only if statements in the program, its time complexity is O(1).
• Auxiliary Space: O(1)
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