Find the day number in the current year for the given date

Given string str which represents a date formatted as YYYY-MM-DD, the task is to find the day number for the current year. For example, 1^st January is the 1^st day of the year, 2^nd January is the 2^nd day of the year, 1^st February is the 32^nd day of the year and so on.

Examples:

Input: str = “2019-01-09” 
Output: 9

Input: str = “2003-03-01” 
Output: 60 

Approach:

• Extract the year, month and the day from the given date and store them in variables year, month and day.
• Create an array days[] where days[i] will store the number of days in the i^th month.
• Update count = days[0] + days[1] + … + days[month – 1] to get the count of all the past days of previous months.
• If the given year is a leap year then increment this count by 1 in order to count 29^th February.
• Finally, add day to the count which is the number of the day in the current month and print the final count.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

int days[] = { 31, 28, 31, 30, 31, 30,
               31, 31, 30, 31, 30, 31 };

// Function to return the day number
// of the year for the given date
int dayOfYear(string date)
{
    // Extract the year, month and the
    // day from the date string
    int year = stoi(date.substr(0, 4));
    int month = stoi(date.substr(5, 2));
    int day = stoi(date.substr(8));

    // If current year is a leap year and the date
    // given is after the 28th of February then
    // it must include the 29th February
    if (month > 2 && year % 4 == 0
        && (year % 100 != 0 || year % 400 == 0)) {
        ++day;
    }

    // Add the days in the previous months
    while (month-- > 0) {
        day = day + days[month - 1];
    }
    return day;
}

// Driver code
int main()
{
    string date = "2019-01-09";
    cout << dayOfYear(date);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{
    static int days [] = { 31, 28, 31, 30, 31, 30,
                           31, 31, 30, 31, 30, 31 };

    // Function to return the day number
    // of the year for the given date
    static int dayOfYear(String date)
    {
        // Extract the year, month and the
        // day from the date string
        int year = Integer.parseInt(date.substring(0, 4));

        int month = Integer.parseInt(date.substring(5, 7));

        int day = Integer.parseInt(date.substring(8));

        // If current year is a leap year and the date
        // given is after the 28th of February then
        // it must include the 29th February
        if (month > 2 && year % 4 == 0 &&
           (year % 100 != 0 || year % 400 == 0))
        {
            ++day;
        }

        // Add the days in the previous months
        while (--month > 0)
        {
            day = day + days[month - 1];
        }
        return day;
    }

    // Driver code
    public static void main (String[] args)
    {
        String date = "2019-01-09";
        System.out.println(dayOfYear(date));
    }
}

// This code is contributed by ihritik

Python3

# Python3 implementation of the approach
days = [31, 28, 31, 30, 31, 30,
        31, 31, 30, 31, 30, 31];

# Function to return the day number
# of the year for the given date
def dayOfYear(date):

    # Extract the year, month and the
    # day from the date string
    year = (int)(date[0:4]);
    month = (int)(date[5:7]);
    day = (int)(date[8:]);

    # If current year is a leap year and the date
    # given is after the 28th of February then
    # it must include the 29th February
    if (month > 2 and year % 4 == 0 and
       (year % 100 != 0 or year % 400 == 0)):
        day += 1;

    # Add the days in the previous months
    month -= 1;
    while (month > 0):
        day = day + days[month - 1];
        month -= 1;
    return day;

# Driver code
if __name__ == '__main__':
    date = "2019-01-09";
    print(dayOfYear(date));

# This code is contributed by Rajput-Ji

C#

// C# implementation of the approach
using System;

class GFG
{
    static int [] days = { 31, 28, 31, 30, 31, 30,
                           31, 31, 30, 31, 30, 31 };

    // Function to return the day number
    // of the year for the given date
    static int dayOfYear(string date)
    {
        // Extract the year, month and the
        // day from the date string
        int year = Int32.Parse(date.Substring(0, 4));

        int month = Int32.Parse(date.Substring(5, 2));

        int day = Int32.Parse(date.Substring(8));

        // If current year is a leap year and the date
        // given is after the 28th of February then
        // it must include the 29th February
        if (month > 2 && year % 4 == 0 &&
           (year % 100 != 0 || year % 400 == 0))
        {
            ++day;
        }

        // Add the days in the previous months
        while (--month > 0)
        {
            day = day + days[month - 1];

        }
        return day;
    }

    // Driver code
    public static void Main ()
    {
        String date = "2019-01-09";
        Console.WriteLine(dayOfYear(date));
    }
}

// This code is contributed by ihritik

Javascript

<script>

// Javascript implementation of the approach

var days= [31, 28, 31, 30, 31, 30,
               31, 31, 30, 31, 30, 31 ];

// Function to return the day number
// of the year for the given date
function dayOfYear( date)
{
    // Extract the year, month and the
    // day from the date string
    var year = parseInt(date.substring(0, 4));
    var month = parseInt(date.substring(5, 6));
    var day = parseInt(date.substring(8));

    // If current year is a leap year and the date
    // given is after the 28th of February then
    // it must include the 29th February
    if (month > 2 && year % 4 == 0
        && (year % 100 != 0 || year % 400 == 0)) {
        ++day;
    }

    // Add the days in the previous months
    while (month-- > 0) {
        day = day + days[month - 1];
    }
    return day;
}

// Driver code
var date = "2019-01-09";
document.write( dayOfYear(date));

// This code is contributed by rutvik_56.
</script>

Output

9

Time Complexity: O(1)
Auxiliary Space: O(1)