# Clear solution by finding overlapped length on x/y

• In general we just need to find out the overlapped rect area (if it exists), so we need to find out the overlapped length on x and y directions.

1. check the 2 rects and find which one is on the left, name it r1 (the one on the right is r2)
2. now check the right boundary. there are only 2 possibilities: r2 is within r1 (on x axis only) or r2/r1 is overlapping. Then we can get the overlapped length on x
3. do similar actions for y directions to find the overlapped length on y

result = r1.size() + r2.size() - overlapped_x*overlapped_y;

The key point is when we check for x-axis, forget about y, and vice versa.

``````class Solution {
struct Rect {
int x1;
int y1;
int x2;
int y2;
int size() {
return (x2-x1)*(y2-y1);
}
Rect(int a,int b,int c, int d):x1(a),y1(b),x2(c),y2(d) {}
};

public:
int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
Rect r1(A,B,C,D);
Rect r2(E,F,G,H);
if (A>E) {
Rect temp = r2;
r2=r1;
r1=temp;
}

if (r1.x2 > r2.x1) {
int ox = r2.x2 < r1.x2 ? (r2.x2-r2.x1) : (r1.x2-r2.x1);  //overlapped x

if (r1.y1 > r2.y1) {
Rect temp = r1;
r1 = r2;
r2 = temp;
}
// r1 is lower
if (r1.y2 > r2.y1) {
int oy = r2.y2 < r1.y2 ? (r2.y2 - r2.y1) : (r1.y2 - r2.y1);
return r1.size()+r2.size() - ox*oy;
}
}

return r1.size() + r2.size();
}
};``````

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