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

Given a 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, 1st January is the 1st day of the year, 2nd January is the 2nd day of the year, 1st February is the 32nd day of the year and so on.

Examples:

Input: str = “2019-01-09”
Output: 9

Input: str = “2003-03-01”
Output: 60

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ith 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 29th February.
• Finally, add day to the count which is 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 ` `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 `

Output:

```9
```

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Improved By : ihritik, Rajput-Ji

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