<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 rectangleArea(vector<vector<int>>& rectangles) {
        int n = rectangles.size();
        sort(rectangles.begin(), rectangles.end(), [](auto& a, auto& b) {
            return a[1] < b[1];
        });

        vector<int> points;
        for (int i = 0; i < n; i++) {
            points.push_back(rectangles[i][0]);
            points.push_back(rectangles[i][2]);
        }
        sort(points.begin(), points.end());
        points.erase(unique(points.begin(), points.end()), points.end());

        long long ans = 0;
        for (int i = 1; i < points.size(); i++) {
            int currL = 0;
            int l = -1;
            int r = -1;
            for (int j = 0; j < n; j++) {
                auto& rect = rectangles[j];
                if (rect[2] <= points[i - 1]) continue;
                if (rect[0] >= points[i]) continue;
                if (rect[1] > r) {
                    currL += r - l;
                    l = rect[1];
                    r = rect[3];
                } else {
                    r = max(r, rect[3]);
                }
            }
            currL += r - l;
            ans += 1LL * currL * (points[i] - points[i - 1]);
        }

        return (int) (ans % 1000000007);
    }
};

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