<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> normDays = {0,31,28,31,30,31,30,31,31,30,31,30,31};
        vector<int> leapDays = {0,31,29,31,30,31,30,31,31,30,31,30,31};

        for(int i = 1; i<normDays.size();i++){
            normDays[i] += normDays[i-1];
            leapDays[i] += leapDays[i-1];
        }

        int y = stoi(date.substr(0,4));
        int m = stoi(date.substr(5,2));
        int d = stoi(date.substr(8,2));
        int ans = -1;

        if(y%400 == 0 || (y%4==0 && y%100 != 0)){
            ans = leapDays[m-1] + d;
        }else{
            ans = normDays[m-1]+d;
        }

        return ans;
    }
};
13400