<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 {
    array<int, 12> days_per_month {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    bool is_leap(int year)
    {
        if (year % 400 == 0)
        {
            return true;
        }
        if (year % 4 == 0 && year % 100 != 0)
        {
            return true;
        }
        return false;
    }
public:
    int dayOfYear(string date) {
        for (auto& c : date)
        {
            if (c == '-')
            {
                c = ' ';
            }
        }
        int year;
        int month;
        int day;
        istringstream is(date);
        is >> year >> month >> day;
        if (is_leap(year))
        {
            days_per_month[1] += 1;
        }
        return accumulate(days_per_month.begin(), days_per_month.begin() + month - 1, 0) + day;
    }
};
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