<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) {
        set<int> x;
        for(auto r:rectangles) {
            x.insert(r[0]);
            x.insert(r[2]);
        }
        vector<int> xc;
        for(auto e:x) xc.push_back(e);

        vector<vector<vector<int>>> range(xc.size()-1);
        for(int i=0;i<range.size();i++) {
            int t = xc[i];
            for(auto r:rectangles) {
                if(r[0] <= t && r[2] > t) {
                    range[i].push_back({r[1], r[3]});
                }
            }
        }

        for(auto &r:range)
            sort(r.begin(), r.end());

        long long N = 1e9+7, ans = 0;

        for(int i=0;i<range.size();i++) {
            long long w = xc[i+1] - xc[i];
            long long l = helper(range[i]);

            ans = (ans + l*w) % N;
        }

        return ans;
    }

    long long helper(vector<vector<int>>& v) {
        if(v.empty()) return 0;
        vector<vector<int>> t = merge(v);

        long long ans = 0;
        for(auto e:t) {
            ans += e[1] - e[0];
        }

        return ans;
    }

    vector<vector<int>> merge(vector<vector<int>>& arr) {
        vector<vector<int>> ans;
	    ans.push_back(arr[0]);
	    for(auto a:arr) {
		    if(a[0]<=ans[ans.size()-1][1])
			    ans[ans.size()-1][1] = max(a[1],ans[ans.size()-1][1]);
		    else
			    ans.push_back(a);
	    }
	    return ans;
    }
};
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