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 )
        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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