<p>You are given a 2D array of axis-aligned <code>rectangles</code>. Each <code>rectangle[i] = [x<sub>i1</sub>, y<sub>i1</sub>, x<sub>i2</sub>, y<sub>i2</sub>]</code> denotes the <code>i<sup>th</sup></code> rectangle where <code>(x<sub>i1</sub>, y<sub>i1</sub>)</code> are the coordinates of the <strong>bottom-left corner</strong>, and <code>(x<sub>i2</sub>, y<sub>i2</sub>)</code> are the coordinates of the <strong>top-right corner</strong>.</p>

<p>Calculate the <strong>total area</strong> covered by all <code>rectangles</code> in the plane. Any area covered by two or more rectangles should only be counted <strong>once</strong>.</p>

<p>Return <em>the <strong>total area</strong></em>. Since the answer may be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://s3-lc-upload.s3.amazonaws.com/uploads/2018/06/06/rectangle_area_ii_pic.png" style="width: 600px; height: 450px;" />
<pre>
<strong>Input:</strong> rectangles = [[0,0,2,2],[1,0,2,3],[1,0,3,1]]
<strong>Output:</strong> 6
<strong>Explanation:</strong> A total area of 6 is covered by all three rectangles, as illustrated in the picture.
From (1,1) to (2,2), the green and red rectangles overlap.
From (1,0) to (2,3), all three rectangles overlap.
</pre>

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

<pre>
<strong>Input:</strong> rectangles = [[0,0,1000000000,1000000000]]
<strong>Output:</strong> 49
<strong>Explanation:</strong> The answer is 10<sup>18</sup> modulo (10<sup>9</sup> + 7), which is 49.
</pre>

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

<ul>
	<li><code>1 &lt;= rectangles.length &lt;= 200</code></li>
	<li><code>rectanges[i].length == 4</code></li>
	<li><code>0 &lt;= x<sub>i1</sub>, y<sub>i1</sub>, x<sub>i2</sub>, y<sub>i2</sub> &lt;= 10<sup>9</sup></code></li>
	<li><code>x<sub>i1 &lt;= </sub>x<sub>i2</sub></code></li>
	<li><code>y<sub>i1 &lt;=</sub> y<sub>i2</sub></code></li>
</ul>

class Solution {
public:

    int MOD = 10000'00007;

    long long merge(multiset<pair<int,int>> intervals)
    {
        vector<pair<int,int>> merged;

        for(auto& ele : intervals)
        {
            if(merged.empty())
            {
                merged.push_back(ele);
            }
            else
            {
                int end = merged.back().second;
                if(ele.first <= end)
                {
                    merged.back().second = max(end, ele.second);
                }
                else
                {
                    merged.push_back(ele);
                }
            }
        }

        long long height = 0;

        for(int i = 0; i<merged.size(); i++)
        {
            int len = merged[i].second - merged[i].first;
            height = (height + (long long)len)%MOD;
        }

        return height;
    }

    int rectangleArea(vector<vector<int>>& rectangles) {
        // sweep line after sorting by x coordinate, add and remove y lines
        // https://leetcode.com/problems/rectangle-area-ii/solutions/1419181/python-4-solutions-n-3-n-2-log-n-n-2-n-log-n-explained

        vector<vector<int>> sides;

        for(int i = 0; i<rectangles.size(); i++)
        {
            int x1 = rectangles[i][0];
            int y1 = rectangles[i][1];
            int x2 = rectangles[i][2];
            int y2 = rectangles[i][3];
            vector<int> a = {x1, 0, y1, y2};
            vector<int> b = {x2, 1, y1, y2};
            sides.push_back(a);
            sides.push_back(b);
        }

        sort(sides.begin(), sides.end());

        long long prev_x = sides[0][0];
        long long area = 0;
        multiset<pair<int,int>> intervals;

        for(int i = 0; i<sides.size(); i++)
        {
            long long x = sides[i][0];
            int op = sides[i][1];
            int y1 = sides[i][2];
            int y2 = sides[i][3];
            area = (area + (merge(intervals) * (x - prev_x))%MOD)%MOD;

            if(op == 0)
            {
                intervals.insert({y1, y2});
            }
            else
            {
                intervals.erase(intervals.find({y1, y2}));
            }

            prev_x = x;
        }

        return (int)area;

    }
};
15200