Program to count the number of days between two years

Given two years. Write a program to count the number of days between the two years.

Examples:

Input: startYear = 1990, endYear = 2001
Output: 4383
Explanation: Years 1992, 1996 and 2000 are leap year and all 9 others are non-leap years. So, total number of days are: 366 * 3 + 365 * 9 = 4383

Input: startYear = 1890, endYear = 1998
Output: 39811

Approach: To solve the problem, follow the below idea:

The Approach is to iterate from the starting year to the ending year, and for each of them check if the given year is a leap year or not, as leap year has 366 days while non leap year has 365 days.

Step-by-step approach:

• Iterate from startYear to endYear.
• For Each iteration, check for leap year or non-leap year.
• If year is a leap year add 366 days in the noOfDays var.
• Else add 365 days in the noOfDays.

Below is the implementation of the above approach:

C++

#include <iostream>;
using namespace std;

bool isLeapYear(int year)
{
    // Leap year is divisible by 4, but not divisible by 100
    // unless divisible by 400
    return (year % 4 == 0 && year % 100 != 0)
           || (year % 400 == 0);
}

int daysInYear(int year)
{
    if (isLeapYear(year))
        return 366;
    return 365;
}

int main()
{
    int startYear = 1890, endYear = 1998;

    // Calculate the number of days between the two years
    int numberOfDays = 0;
    for (int year = startYear; year <= endYear; ++year) {
        numberOfDays += daysInYear(year);
    }

    // Display the result
    cout << "Number of days between " << startYear
         << " and " << endYear << " is: " << numberOfDays
         << " days." << endl;

    return 0;
}

Java

public class NumberOfDaysCalculator {
    public static boolean isLeapYear(int year)
    {
        // Leap year is divisible by 4, but not divisible by
        // 100 unless divisible by 400
        return (year % 4 == 0 && year % 100 != 0)
            || (year % 400 == 0);
    }

    public static int daysInYear(int year)
    {
        return isLeapYear(year) ? 366 : 365;
    }

    public static void main(String[] args)
    {
        int startYear = 1890, endYear = 1998;

        // Calculate the number of days between the two
        // years
        int numberOfDays = 0;
        for (int year = startYear; year <= endYear;
             ++year) {
            numberOfDays += daysInYear(year);
        }

        // Display the result
        System.out.println(
            "Number of days between " + startYear + " and "
            + endYear + " is: " + numberOfDays + " days.");
    }
}

Python3

def is_leap_year(year):
    # Leap year is divisible by 4, but not divisible by 100 unless divisible by 400
    return (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0)


def days_in_year(year):
    return 366 if is_leap_year(year) else 365


def main():
    start_year, end_year = 1890, 1998

    # Calculate the number of days between the two years
    number_of_days = sum(days_in_year(year)
                         for year in range(start_year, end_year + 1))

    # Display the result
    print(
        f"Number of days between {start_year} and {end_year} is: {number_of_days} days.")


if __name__ == "__main__":
    main()

C#

using System;

class Program {
    static bool IsLeapYear(int year)
    {
        // Leap year is divisible by 4, but not divisible by
        // 100 unless divisible by 400
        return (year % 4 == 0 && year % 100 != 0)
            || (year % 400 == 0);
    }

    static int DaysInYear(int year)
    {
        return IsLeapYear(year) ? 366 : 365;
    }

    static void Main()
    {
        int startYear = 1890, endYear = 1998;

        // Calculate the number of days between the two
        // years
        int numberOfDays = 0;
        for (int year = startYear; year <= endYear;
             ++year) {
            numberOfDays += DaysInYear(year);
        }

        // Display the result
        Console.WriteLine(
            $"Number of days between {startYear} and {endYear} is: {numberOfDays} days.");
    }
}

Javascript

function isLeapYear(year) {
    // Leap year is divisible by 4, but not divisible by 100 unless divisible by 400
    return (year % 4 === 0 && year % 100 !== 0) || (year % 400 === 0);
}

function daysInYear(year) {
    return isLeapYear(year) ? 366 : 365;
}

function main() {
    let startYear = 1890, endYear = 1998;

    // Calculate the number of days between the two years
    let numberOfDays = 0;
    for (let year = startYear; year <= endYear; ++year) {
        numberOfDays += daysInYear(year);
    }

    // Display the result
    console.log(`Number of days between ${startYear} and ${endYear} is: ${numberOfDays} days.`);
}

main();

Output

Number of days between 1890 and 1998 is: 39811 days.

Time Complexity: O(N), where N is the number of years between start and end year.
Auxiliary space: O(1)