Find the intersect area of two overlapping rectangles
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This is my solution to find the coordinates of 2 overlapped rectangles implemented in JavaScript. Each rectangle is represented by 2 points, each with 2 (x,y) coordinates.
Can this code be improved?
const maxOfX = (rec) => (Math.max(rec.x1, rec.x2));
const maxOfY = (rec) => (Math.max(rec.y1, rec.y2));
const minOfX = (rec) => (Math.min(rec.x1, rec.x2));
const minOfY = (rec) => (Math.min(rec.y1, rec.y2));
const comBinedRectangle = (rec1, rec2) =>
const rectangle1 = x1: 2, y1: 2, x2: 4, y2: 4 ;
const rectangle2 = x1: 3, y1: 3, x2: 6, y2: 2 ;
console.log(comBinedRectangle(rectangle1, rectangle2));
// x1: 3, y1: 2, x2: 4, y2: 3
javascript performance algorithm interview-questions collision
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up vote
2
down vote
favorite
This is my solution to find the coordinates of 2 overlapped rectangles implemented in JavaScript. Each rectangle is represented by 2 points, each with 2 (x,y) coordinates.
Can this code be improved?
const maxOfX = (rec) => (Math.max(rec.x1, rec.x2));
const maxOfY = (rec) => (Math.max(rec.y1, rec.y2));
const minOfX = (rec) => (Math.min(rec.x1, rec.x2));
const minOfY = (rec) => (Math.min(rec.y1, rec.y2));
const comBinedRectangle = (rec1, rec2) =>
const rectangle1 = x1: 2, y1: 2, x2: 4, y2: 4 ;
const rectangle2 = x1: 3, y1: 3, x2: 6, y2: 2 ;
console.log(comBinedRectangle(rectangle1, rectangle2));
// x1: 3, y1: 2, x2: 4, y2: 3
javascript performance algorithm interview-questions collision
bumped to the homepage by Community⦠yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Don't write, never present undocumented/uncommented code. Don't do(boolean expression) ? true : false
. Try to limit line length.
â greybeard
Jan 17 at 15:44
Are you talking about combination, intersection or even bounding rectangles? Why put 'overlapping' in the name when non-intersecting rectangles are equally valid input?
â le_m
Jan 17 at 17:43
Why not create a Rectange class? I did something similar whilst working on a visualisation project and it helped make sense of the code e.g.rect1.intersects(rect2)
â James
Jan 20 at 0:19
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This is my solution to find the coordinates of 2 overlapped rectangles implemented in JavaScript. Each rectangle is represented by 2 points, each with 2 (x,y) coordinates.
Can this code be improved?
const maxOfX = (rec) => (Math.max(rec.x1, rec.x2));
const maxOfY = (rec) => (Math.max(rec.y1, rec.y2));
const minOfX = (rec) => (Math.min(rec.x1, rec.x2));
const minOfY = (rec) => (Math.min(rec.y1, rec.y2));
const comBinedRectangle = (rec1, rec2) =>
const rectangle1 = x1: 2, y1: 2, x2: 4, y2: 4 ;
const rectangle2 = x1: 3, y1: 3, x2: 6, y2: 2 ;
console.log(comBinedRectangle(rectangle1, rectangle2));
// x1: 3, y1: 2, x2: 4, y2: 3
javascript performance algorithm interview-questions collision
This is my solution to find the coordinates of 2 overlapped rectangles implemented in JavaScript. Each rectangle is represented by 2 points, each with 2 (x,y) coordinates.
Can this code be improved?
const maxOfX = (rec) => (Math.max(rec.x1, rec.x2));
const maxOfY = (rec) => (Math.max(rec.y1, rec.y2));
const minOfX = (rec) => (Math.min(rec.x1, rec.x2));
const minOfY = (rec) => (Math.min(rec.y1, rec.y2));
const comBinedRectangle = (rec1, rec2) =>
const rectangle1 = x1: 2, y1: 2, x2: 4, y2: 4 ;
const rectangle2 = x1: 3, y1: 3, x2: 6, y2: 2 ;
console.log(comBinedRectangle(rectangle1, rectangle2));
// x1: 3, y1: 2, x2: 4, y2: 3
javascript performance algorithm interview-questions collision
edited Jun 19 at 10:29
Przemek
1,032213
1,032213
asked Jan 17 at 15:33
Achref Boukihili
112
112
bumped to the homepage by Community⦠yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community⦠yesterday
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Don't write, never present undocumented/uncommented code. Don't do(boolean expression) ? true : false
. Try to limit line length.
â greybeard
Jan 17 at 15:44
Are you talking about combination, intersection or even bounding rectangles? Why put 'overlapping' in the name when non-intersecting rectangles are equally valid input?
â le_m
Jan 17 at 17:43
Why not create a Rectange class? I did something similar whilst working on a visualisation project and it helped make sense of the code e.g.rect1.intersects(rect2)
â James
Jan 20 at 0:19
add a comment |Â
Don't write, never present undocumented/uncommented code. Don't do(boolean expression) ? true : false
. Try to limit line length.
â greybeard
Jan 17 at 15:44
Are you talking about combination, intersection or even bounding rectangles? Why put 'overlapping' in the name when non-intersecting rectangles are equally valid input?
â le_m
Jan 17 at 17:43
Why not create a Rectange class? I did something similar whilst working on a visualisation project and it helped make sense of the code e.g.rect1.intersects(rect2)
â James
Jan 20 at 0:19
Don't write, never present undocumented/uncommented code. Don't do
(boolean expression) ? true : false
. Try to limit line length.â greybeard
Jan 17 at 15:44
Don't write, never present undocumented/uncommented code. Don't do
(boolean expression) ? true : false
. Try to limit line length.â greybeard
Jan 17 at 15:44
Are you talking about combination, intersection or even bounding rectangles? Why put 'overlapping' in the name when non-intersecting rectangles are equally valid input?
â le_m
Jan 17 at 17:43
Are you talking about combination, intersection or even bounding rectangles? Why put 'overlapping' in the name when non-intersecting rectangles are equally valid input?
â le_m
Jan 17 at 17:43
Why not create a Rectange class? I did something similar whilst working on a visualisation project and it helped make sense of the code e.g.
rect1.intersects(rect2)
â James
Jan 20 at 0:19
Why not create a Rectange class? I did something similar whilst working on a visualisation project and it helped make sense of the code e.g.
rect1.intersects(rect2)
â James
Jan 20 at 0:19
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
- Document/comment
I have no idea whether there are agreed standards for ECMAScript, e.g. JSDoc.
If I used what you calledOverX
I'd call itOverlapX
and invert it.OverlapX = minOfX(rec1) <= maxOfX(rec2) && minOfX(rec2) <= maxOfX(rec1);
- precomputing booleans the way of
OverX
andOverY
precludes short-circuiting the evaluation of the combined no-overlap-condition. - One could use "nomalised" rects:
minx
/miny
/maxx
/maxy
. - (I'd prefer plain function definitions over arrow functions if they are going to get a name, anyway.)
- (There is bound to be a way to have, e.g.
Math.min(values)
operate on all x-coordinates of a shape, my ignorance thereof notwithstanding.)
add a comment |Â
up vote
0
down vote
/**
* Returns intersecting part of two rectangles
* @param object r1 4 coordinates in form of x1, y1, x2, y2 object
* @param object r2 4 coordinates in form of x1, y1, x2, y2 object
* @return boolean False if there's no intersecting part
* @return object 4 coordinates in form of x1, y1, x2, y2 object
*/
const getIntersectingRectangle = (r1, r2) => r2.y[1] < r1.y[0];
return noIntersect ? false :
x1: Math.max(r1.x[0], r2.x[0]), // _[0] is the lesser,
y1: Math.max(r1.y[0], r2.y[0]), // _[1] is the greater
x2: Math.min(r1.x[1], r2.x[1]),
y2: Math.min(r1.y[1], r2.y[1])
;
;
/* â DEMO â */
const rectangle1 = x1: 2, y1: 2, x2: 4, y2: 4 ;
const rectangle2 = x1: 3, y1: 3, x2: 6, y2: 2 ;
console.log(getIntersectingRectangle(rectangle1, rectangle2));
// x1: 3, y1: 2, x2: 4, y2: 3
First, both rectangles get transformed into object with keys x
and y
and sorted arrays of two of corresponding them coordinates as values.
r1 =
x: [2, 4], // x1, x2
y: [2, 4] // y1, y2
r2 =
x: [3, 6], // x1, x2
y: [2, 3] // Y2, Y1 !
That's because for this job it is important to know which of x
s and y
s is lesser and which is greater. Rather than Math.max()
and Math.min()
a single .sort()
can be used.
Alternative would be to declare that (x1, y1)
is assumed to always be top-left corner, but as I see in rectangle2
in your question, that is apparently not always the case.
Now, noIntersect
is negated
!(a.left > b.right || b.left > a.right || a.top > b.bottom || b.top > a.bottom);
which tests if in both axes sides from one end of one figure exceed opposite-direction end side of the other figure, e.g. if left side of figure a
is more to the right than the right side of figure b
.
At the end we return false if there is no intersection of our rectangles, or an object with coordinates if there is. Intersecting part will always span from:
the greater of the 2 lesser x
s of both rectangles
to
the lesser of the 2 greater x
s of both rectangles.
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
- Document/comment
I have no idea whether there are agreed standards for ECMAScript, e.g. JSDoc.
If I used what you calledOverX
I'd call itOverlapX
and invert it.OverlapX = minOfX(rec1) <= maxOfX(rec2) && minOfX(rec2) <= maxOfX(rec1);
- precomputing booleans the way of
OverX
andOverY
precludes short-circuiting the evaluation of the combined no-overlap-condition. - One could use "nomalised" rects:
minx
/miny
/maxx
/maxy
. - (I'd prefer plain function definitions over arrow functions if they are going to get a name, anyway.)
- (There is bound to be a way to have, e.g.
Math.min(values)
operate on all x-coordinates of a shape, my ignorance thereof notwithstanding.)
add a comment |Â
up vote
0
down vote
- Document/comment
I have no idea whether there are agreed standards for ECMAScript, e.g. JSDoc.
If I used what you calledOverX
I'd call itOverlapX
and invert it.OverlapX = minOfX(rec1) <= maxOfX(rec2) && minOfX(rec2) <= maxOfX(rec1);
- precomputing booleans the way of
OverX
andOverY
precludes short-circuiting the evaluation of the combined no-overlap-condition. - One could use "nomalised" rects:
minx
/miny
/maxx
/maxy
. - (I'd prefer plain function definitions over arrow functions if they are going to get a name, anyway.)
- (There is bound to be a way to have, e.g.
Math.min(values)
operate on all x-coordinates of a shape, my ignorance thereof notwithstanding.)
add a comment |Â
up vote
0
down vote
up vote
0
down vote
- Document/comment
I have no idea whether there are agreed standards for ECMAScript, e.g. JSDoc.
If I used what you calledOverX
I'd call itOverlapX
and invert it.OverlapX = minOfX(rec1) <= maxOfX(rec2) && minOfX(rec2) <= maxOfX(rec1);
- precomputing booleans the way of
OverX
andOverY
precludes short-circuiting the evaluation of the combined no-overlap-condition. - One could use "nomalised" rects:
minx
/miny
/maxx
/maxy
. - (I'd prefer plain function definitions over arrow functions if they are going to get a name, anyway.)
- (There is bound to be a way to have, e.g.
Math.min(values)
operate on all x-coordinates of a shape, my ignorance thereof notwithstanding.)
- Document/comment
I have no idea whether there are agreed standards for ECMAScript, e.g. JSDoc.
If I used what you calledOverX
I'd call itOverlapX
and invert it.OverlapX = minOfX(rec1) <= maxOfX(rec2) && minOfX(rec2) <= maxOfX(rec1);
- precomputing booleans the way of
OverX
andOverY
precludes short-circuiting the evaluation of the combined no-overlap-condition. - One could use "nomalised" rects:
minx
/miny
/maxx
/maxy
. - (I'd prefer plain function definitions over arrow functions if they are going to get a name, anyway.)
- (There is bound to be a way to have, e.g.
Math.min(values)
operate on all x-coordinates of a shape, my ignorance thereof notwithstanding.)
edited Jan 17 at 16:38
answered Jan 17 at 16:09
greybeard
1,3231521
1,3231521
add a comment |Â
add a comment |Â
up vote
0
down vote
/**
* Returns intersecting part of two rectangles
* @param object r1 4 coordinates in form of x1, y1, x2, y2 object
* @param object r2 4 coordinates in form of x1, y1, x2, y2 object
* @return boolean False if there's no intersecting part
* @return object 4 coordinates in form of x1, y1, x2, y2 object
*/
const getIntersectingRectangle = (r1, r2) => r2.y[1] < r1.y[0];
return noIntersect ? false :
x1: Math.max(r1.x[0], r2.x[0]), // _[0] is the lesser,
y1: Math.max(r1.y[0], r2.y[0]), // _[1] is the greater
x2: Math.min(r1.x[1], r2.x[1]),
y2: Math.min(r1.y[1], r2.y[1])
;
;
/* â DEMO â */
const rectangle1 = x1: 2, y1: 2, x2: 4, y2: 4 ;
const rectangle2 = x1: 3, y1: 3, x2: 6, y2: 2 ;
console.log(getIntersectingRectangle(rectangle1, rectangle2));
// x1: 3, y1: 2, x2: 4, y2: 3
First, both rectangles get transformed into object with keys x
and y
and sorted arrays of two of corresponding them coordinates as values.
r1 =
x: [2, 4], // x1, x2
y: [2, 4] // y1, y2
r2 =
x: [3, 6], // x1, x2
y: [2, 3] // Y2, Y1 !
That's because for this job it is important to know which of x
s and y
s is lesser and which is greater. Rather than Math.max()
and Math.min()
a single .sort()
can be used.
Alternative would be to declare that (x1, y1)
is assumed to always be top-left corner, but as I see in rectangle2
in your question, that is apparently not always the case.
Now, noIntersect
is negated
!(a.left > b.right || b.left > a.right || a.top > b.bottom || b.top > a.bottom);
which tests if in both axes sides from one end of one figure exceed opposite-direction end side of the other figure, e.g. if left side of figure a
is more to the right than the right side of figure b
.
At the end we return false if there is no intersection of our rectangles, or an object with coordinates if there is. Intersecting part will always span from:
the greater of the 2 lesser x
s of both rectangles
to
the lesser of the 2 greater x
s of both rectangles.
add a comment |Â
up vote
0
down vote
/**
* Returns intersecting part of two rectangles
* @param object r1 4 coordinates in form of x1, y1, x2, y2 object
* @param object r2 4 coordinates in form of x1, y1, x2, y2 object
* @return boolean False if there's no intersecting part
* @return object 4 coordinates in form of x1, y1, x2, y2 object
*/
const getIntersectingRectangle = (r1, r2) => r2.y[1] < r1.y[0];
return noIntersect ? false :
x1: Math.max(r1.x[0], r2.x[0]), // _[0] is the lesser,
y1: Math.max(r1.y[0], r2.y[0]), // _[1] is the greater
x2: Math.min(r1.x[1], r2.x[1]),
y2: Math.min(r1.y[1], r2.y[1])
;
;
/* â DEMO â */
const rectangle1 = x1: 2, y1: 2, x2: 4, y2: 4 ;
const rectangle2 = x1: 3, y1: 3, x2: 6, y2: 2 ;
console.log(getIntersectingRectangle(rectangle1, rectangle2));
// x1: 3, y1: 2, x2: 4, y2: 3
First, both rectangles get transformed into object with keys x
and y
and sorted arrays of two of corresponding them coordinates as values.
r1 =
x: [2, 4], // x1, x2
y: [2, 4] // y1, y2
r2 =
x: [3, 6], // x1, x2
y: [2, 3] // Y2, Y1 !
That's because for this job it is important to know which of x
s and y
s is lesser and which is greater. Rather than Math.max()
and Math.min()
a single .sort()
can be used.
Alternative would be to declare that (x1, y1)
is assumed to always be top-left corner, but as I see in rectangle2
in your question, that is apparently not always the case.
Now, noIntersect
is negated
!(a.left > b.right || b.left > a.right || a.top > b.bottom || b.top > a.bottom);
which tests if in both axes sides from one end of one figure exceed opposite-direction end side of the other figure, e.g. if left side of figure a
is more to the right than the right side of figure b
.
At the end we return false if there is no intersection of our rectangles, or an object with coordinates if there is. Intersecting part will always span from:
the greater of the 2 lesser x
s of both rectangles
to
the lesser of the 2 greater x
s of both rectangles.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
/**
* Returns intersecting part of two rectangles
* @param object r1 4 coordinates in form of x1, y1, x2, y2 object
* @param object r2 4 coordinates in form of x1, y1, x2, y2 object
* @return boolean False if there's no intersecting part
* @return object 4 coordinates in form of x1, y1, x2, y2 object
*/
const getIntersectingRectangle = (r1, r2) => r2.y[1] < r1.y[0];
return noIntersect ? false :
x1: Math.max(r1.x[0], r2.x[0]), // _[0] is the lesser,
y1: Math.max(r1.y[0], r2.y[0]), // _[1] is the greater
x2: Math.min(r1.x[1], r2.x[1]),
y2: Math.min(r1.y[1], r2.y[1])
;
;
/* â DEMO â */
const rectangle1 = x1: 2, y1: 2, x2: 4, y2: 4 ;
const rectangle2 = x1: 3, y1: 3, x2: 6, y2: 2 ;
console.log(getIntersectingRectangle(rectangle1, rectangle2));
// x1: 3, y1: 2, x2: 4, y2: 3
First, both rectangles get transformed into object with keys x
and y
and sorted arrays of two of corresponding them coordinates as values.
r1 =
x: [2, 4], // x1, x2
y: [2, 4] // y1, y2
r2 =
x: [3, 6], // x1, x2
y: [2, 3] // Y2, Y1 !
That's because for this job it is important to know which of x
s and y
s is lesser and which is greater. Rather than Math.max()
and Math.min()
a single .sort()
can be used.
Alternative would be to declare that (x1, y1)
is assumed to always be top-left corner, but as I see in rectangle2
in your question, that is apparently not always the case.
Now, noIntersect
is negated
!(a.left > b.right || b.left > a.right || a.top > b.bottom || b.top > a.bottom);
which tests if in both axes sides from one end of one figure exceed opposite-direction end side of the other figure, e.g. if left side of figure a
is more to the right than the right side of figure b
.
At the end we return false if there is no intersection of our rectangles, or an object with coordinates if there is. Intersecting part will always span from:
the greater of the 2 lesser x
s of both rectangles
to
the lesser of the 2 greater x
s of both rectangles.
/**
* Returns intersecting part of two rectangles
* @param object r1 4 coordinates in form of x1, y1, x2, y2 object
* @param object r2 4 coordinates in form of x1, y1, x2, y2 object
* @return boolean False if there's no intersecting part
* @return object 4 coordinates in form of x1, y1, x2, y2 object
*/
const getIntersectingRectangle = (r1, r2) => r2.y[1] < r1.y[0];
return noIntersect ? false :
x1: Math.max(r1.x[0], r2.x[0]), // _[0] is the lesser,
y1: Math.max(r1.y[0], r2.y[0]), // _[1] is the greater
x2: Math.min(r1.x[1], r2.x[1]),
y2: Math.min(r1.y[1], r2.y[1])
;
;
/* â DEMO â */
const rectangle1 = x1: 2, y1: 2, x2: 4, y2: 4 ;
const rectangle2 = x1: 3, y1: 3, x2: 6, y2: 2 ;
console.log(getIntersectingRectangle(rectangle1, rectangle2));
// x1: 3, y1: 2, x2: 4, y2: 3
First, both rectangles get transformed into object with keys x
and y
and sorted arrays of two of corresponding them coordinates as values.
r1 =
x: [2, 4], // x1, x2
y: [2, 4] // y1, y2
r2 =
x: [3, 6], // x1, x2
y: [2, 3] // Y2, Y1 !
That's because for this job it is important to know which of x
s and y
s is lesser and which is greater. Rather than Math.max()
and Math.min()
a single .sort()
can be used.
Alternative would be to declare that (x1, y1)
is assumed to always be top-left corner, but as I see in rectangle2
in your question, that is apparently not always the case.
Now, noIntersect
is negated
!(a.left > b.right || b.left > a.right || a.top > b.bottom || b.top > a.bottom);
which tests if in both axes sides from one end of one figure exceed opposite-direction end side of the other figure, e.g. if left side of figure a
is more to the right than the right side of figure b
.
At the end we return false if there is no intersection of our rectangles, or an object with coordinates if there is. Intersecting part will always span from:
the greater of the 2 lesser x
s of both rectangles
to
the lesser of the 2 greater x
s of both rectangles.
/**
* Returns intersecting part of two rectangles
* @param object r1 4 coordinates in form of x1, y1, x2, y2 object
* @param object r2 4 coordinates in form of x1, y1, x2, y2 object
* @return boolean False if there's no intersecting part
* @return object 4 coordinates in form of x1, y1, x2, y2 object
*/
const getIntersectingRectangle = (r1, r2) => r2.y[1] < r1.y[0];
return noIntersect ? false :
x1: Math.max(r1.x[0], r2.x[0]), // _[0] is the lesser,
y1: Math.max(r1.y[0], r2.y[0]), // _[1] is the greater
x2: Math.min(r1.x[1], r2.x[1]),
y2: Math.min(r1.y[1], r2.y[1])
;
;
/* â DEMO â */
const rectangle1 = x1: 2, y1: 2, x2: 4, y2: 4 ;
const rectangle2 = x1: 3, y1: 3, x2: 6, y2: 2 ;
console.log(getIntersectingRectangle(rectangle1, rectangle2));
// x1: 3, y1: 2, x2: 4, y2: 3
/**
* Returns intersecting part of two rectangles
* @param object r1 4 coordinates in form of x1, y1, x2, y2 object
* @param object r2 4 coordinates in form of x1, y1, x2, y2 object
* @return boolean False if there's no intersecting part
* @return object 4 coordinates in form of x1, y1, x2, y2 object
*/
const getIntersectingRectangle = (r1, r2) => r2.y[1] < r1.y[0];
return noIntersect ? false :
x1: Math.max(r1.x[0], r2.x[0]), // _[0] is the lesser,
y1: Math.max(r1.y[0], r2.y[0]), // _[1] is the greater
x2: Math.min(r1.x[1], r2.x[1]),
y2: Math.min(r1.y[1], r2.y[1])
;
;
/* â DEMO â */
const rectangle1 = x1: 2, y1: 2, x2: 4, y2: 4 ;
const rectangle2 = x1: 3, y1: 3, x2: 6, y2: 2 ;
console.log(getIntersectingRectangle(rectangle1, rectangle2));
// x1: 3, y1: 2, x2: 4, y2: 3
answered Jun 19 at 0:22
Przemek
1,032213
1,032213
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Don't write, never present undocumented/uncommented code. Don't do
(boolean expression) ? true : false
. Try to limit line length.â greybeard
Jan 17 at 15:44
Are you talking about combination, intersection or even bounding rectangles? Why put 'overlapping' in the name when non-intersecting rectangles are equally valid input?
â le_m
Jan 17 at 17:43
Why not create a Rectange class? I did something similar whilst working on a visualisation project and it helped make sense of the code e.g.
rect1.intersects(rect2)
â James
Jan 20 at 0:19