<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> thirty{9, 4, 6, 11};
        int year = stoi(date.substr(0, 4));
        int month = stoi(date.substr(5, 2));
        int day = stoi(date.substr(8, 2));
        int num = 0;
        
        for(int i = 1; i <= month; i++){
            if(month == i) { num += day; break; }
            
            if(i == 2){
                if(((2020 - year) % 4 == 0) && year != 1900) num += 29;
                else num += 28;
            }
            else if(count(thirty.begin(), thirty.end(), i)) num += 30;
            else num += 31;
        }
        
        return num;
    }
};
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