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

Leap Year Program in C++

C++

// C++ program to check if a given
// year is a leap year or not
#include <iostream>
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)