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# Rectangle Area LeetCode Solution

## Problem – Rectangle Area LeetCode Solution

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:

• `-104 <= ax1 <= ax2 <= 104`
• `-104 <= ay1 <= ay2 <= 104`
• `-104 <= bx1 <= bx2 <= 104`
• `-104 <= by1 <= by2 <= 104`

## Rectangle Area LeetCode Solution in Java

``````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;
}
``````

## Rectangle Area LeetCode Solution in Python

``````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
``````

## Rectangle Area LeetCode Solution in C++

``````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 )
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 - h ) * (v - v);
}
}
``````
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