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Dorchester Center, MA 02124

Given the coordinates of two **rectilinear** rectangles in a 2D plane, return *the total area covered by the two rectangles*.

The first rectangle is defined by its **bottom-left** corner `(ax1, ay1)`

and its **top-right** corner `(ax2, ay2)`

.

The second rectangle is defined by its **bottom-left** corner `(bx1, by1)`

and its **top-right** corner `(bx2, by2)`

.

**Example 1:**

**Input:** ax1 = -3, ay1 = 0, ax2 = 3, ay2 = 4, bx1 = 0, by1 = -1, bx2 = 9, by2 = 2
**Output:** 45

**Example 2:**

**Input:** ax1 = -2, ay1 = -2, ax2 = 2, ay2 = 2, bx1 = -2, by1 = -2, bx2 = 2, by2 = 2
**Output:** 16

**Constraints:**

`-10`

^{4}<= ax1 <= ax2 <= 10^{4}`-10`

^{4}<= ay1 <= ay2 <= 10^{4}`-10`

^{4}<= bx1 <= bx2 <= 10^{4}`-10`

^{4}<= by1 <= by2 <= 10^{4}

```
public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
int areaOfSqrA = (C-A) * (D-B);
int areaOfSqrB = (G-E) * (H-F);
int left = Math.max(A, E);
int right = Math.min(G, C);
int bottom = Math.max(F, B);
int top = Math.min(D, H);
//If overlap
int overlap = 0;
if(right > left && top > bottom)
overlap = (right - left) * (top - bottom);
return areaOfSqrA + areaOfSqrB - overlap;
}
```

```
def computeArea(self, A, B, C, D, E, F, G, H):
overlap = max(min(C,G)-max(A,E), 0)*max(min(D,H)-max(B,F), 0)
return (A-C)*(B-D) + (E-G)*(F-H) - overlap
```

```
int computeArea(int A, int B, int C, int D, int E, int F, int G, int H)
{
int total = (C-A) * (D-B) + (G-E) * (H-F);
if (C<=E || A>=G || B>=H || D<=F )
return total;
else
{
vector <int> h;
h.push_back(A);
h.push_back(C);
h.push_back(E);
h.push_back(G);
vector <int> v;
v.push_back(B);
v.push_back(D);
v.push_back(F);
v.push_back(H);
sort(h.begin(), h.end());
sort(v.begin(), v.end());
total = total - (h[2] - h [1]) * (v[2] - v[1]);
return total;
}
}
```

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