<p>Given a string <code>date</code> representing a <a href="https://en.wikipedia.org/wiki/Gregorian_calendar" target="_blank">Gregorian calendar</a> date formatted as <code>YYYY-MM-DD</code>, return <em>the day number of the year</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> date = &quot;2019-01-09&quot;
<strong>Output:</strong> 9
<strong>Explanation:</strong> Given date is the 9th day of the year in 2019.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> date = &quot;2019-02-10&quot;
<strong>Output:</strong> 41
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>date.length == 10</code></li>
	<li><code>date[4] == date[7] == &#39;-&#39;</code>, and all other <code>date[i]</code>&#39;s are digits</li>
	<li><code>date</code> represents a calendar date between Jan 1<sup>st</sup>, 1900 and Dec 31<sup>th</sup>, 2019.</li>
</ul>

class Solution {
public:
    int dayOfYear(string date) {
        vector<int> M = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; // February set to 28 by default
        vector<string> D;
        string temp = "";


        for (int i = 0; i < date.size(); i++) {
            if (date[i] != '-') {
                temp = temp + date[i];
            } else {
                D.push_back(temp);
                temp = "";
            }
        }
        D.push_back(temp);  
      
        int y = stoi(D[0]);
        int m = stoi(D[1]);
        int d = stoi(D[2]);

 
        if ((y % 400 == 0) || (y % 4 == 0 && y % 100 != 0)) {
            M[1] = 29;  
        }
        
        
        int sum = 0;
        for (int i = 0; i < m - 1; i++) {  
            sum = sum + M[i];
        }

        return sum + d;  
    }
};

15900