You have a plain with lots of rectangles on it, find out how many of them intersect
https://github.com/allaboutjst/airbnb/blob/master/src/main/java/number_of_intersected_rectangles/NumberofIntersectedRectangles.java
https://www.reddit.com/r/dailyprogrammer/comments/23b1pr/4182014_challenge_158_hard_intersecting_rectangles/
https://www.coursera.org/learn/algorithms-part1/lecture/mNiwq/rectangle-intersection
http://algs4.cs.princeton.edu/93intersection/
http://algs4.cs.princeton.edu/93intersection/Interval1D.java.html
http://algs4.cs.princeton.edu/93intersection/IntervalIntersection.java.html
https://algnotes.wordpress.com/2015/05/20/intersection-of-rectangles/
https://github.com/allaboutjst/airbnb/blob/master/src/main/java/number_of_intersected_rectangles/NumberofIntersectedRectangles.java
private boolean intersect(int[][] r1, int[][] r2) {
return r1[0][0] < r2[0][0] && r1[0][1] < r2[0][1] && r2[0][0] < r1[1][0] && r2[0][1] < r1[1][1] ||
r1[0][0] < r2[1][0] && r1[0][1] < r2[1][1] && r2[1][0] < r1[1][0] && r2[1][1] < r1[1][1];
}
private int find(int val, int[] parents) {
while (parents[val] != val) {
val = parents[val];
}
return val;
}
public int countIntersection(int[][][] rectangles) {
if (rectangles == null || rectangles.length == 0) return 0;
int n = rectangles.length;
int[] parents = new int[n];
for (int i = 0; i < n; i++) {
parents[i] = i;
}
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
if (intersect(rectangles[i], rectangles[j])) {
int root1 = find(i, parents);
int root2 = find(j, parents);
if (root1 != root2) {
parents[root1] = root2;
}
}
}
}
Set<Integer> set = new HashSet<>();
for (int i = 0; i < n; i++) {
set.add(find(i, parents));
}
return set.size();
}
https://www.reddit.com/r/dailyprogrammer/comments/23b1pr/4182014_challenge_158_hard_intersecting_rectangles/
Thinking of each shape individually will only make this challenge harder. Try grouping intersecting shapes up, or calculating the area of regions of the shape at a time.
Allocating occupied points in a 2-D array would be the easy way out of doing this - however, this falls short when you have large shapes, or the points are not integer values. Try to come up with another way of doing it.
Allocating occupied points in a 2-D array would be the easy way out of doing this - however, this falls short when you have large shapes, or the points are not integer values. Try to come up with another way of doing it.
http://algs4.cs.princeton.edu/93intersection/
http://algs4.cs.princeton.edu/93intersection/Interval1D.java.html
http://algs4.cs.princeton.edu/93intersection/IntervalIntersection.java.html
// create events MinPQ<Event> pq = new MinPQ<Event>(); for (int i = 0; i < N; i++) { Event e1 = new Event(intervals[i].min, intervals[i]); Event e2 = new Event(intervals[i].max, intervals[i]); pq.insert(e1); pq.insert(e2); } // run sweep-line algorithm HashSet<Interval1D> st = new HashSet<Interval1D>(); while (!pq.isEmpty()) { Event e = pq.delMin(); int time = e.time; Interval1D interval = e.interval; // next event is the right endpoint of interval i if (time == interval.max) st.remove(interval); // slow // next event is the left endpoint of interval i else { for (Interval1D x : st) { StdOut.println("Intersection: " + interval + ", " + x); } st.add(interval); } }http://www.java-gaming.org/index.php?topic=29567.0
https://algnotes.wordpress.com/2015/05/20/intersection-of-rectangles/
This problem can be solved by using plane-sweep technique. Here, we use divide-and-conquer technique instead using only linear arrays.
We apply divide-and-conquer based on the set of vertical edges. Consider a set of vertical edges, V, of rectangles in increasing x-coordinate. Let P be the first half of V and Q be the second half of V. We classify rectangles according to V into four classes.
- L: the set of rectangles that one vertical edge is in P, and another edge has smaller x-coordinate than any edges in Q.
- L‘: the set of rectangles that one vertical edge is in P, and another edge has larger x-coordinate than any edges in Q.
- R: the set of rectangles that one vertical edge is in Q, and another edge has larger x-coordinate than any edges in P.
- R‘: the set of rectangles that one vertical edge is in Q, and another edge has smaller x-coordinate than any edges in P.
- LR: the set of rectangles that one vertical edge is in P, and another edge is in Q.
The intersections within the same class will be found recursively by recursive call on P and Q respectively. Since the case of LR eventually will become L‘ and R‘ in the recursion, we will ignore the intersection related to LR.
Let’s consider the intersections between different classes. By definition, rectangles in L will not intersect with R. Moreover, the intersections in (L, L‘) and (R, R‘) will be found in later recursion calls. Thus, we only need to consider the intersection in (L’, R), (R’, L), and (L‘, R‘). Since all rectangles in L‘ span the whole range of R in x-coordinate, the intersection between L‘ and R can be checked easily in y-coordinate. The case of (R‘, L) and (L‘, R‘) can be solved similarly.
The total complexity is O(n lg n + k), where k is the number of output. Since each some intersections may be reported twice, we need to remove the duplication in the end of the algorithm.
Frank Dévai and László Neumann, “A Rectangle-Intersection Algorithm with Limited Resource Requirements,” 10th IEEE International Conference on Computer and Information Technology Pages 2335-2340
https://www.careercup.com/question?id=6696054317645824
static class Endpoint implements Comparable<Endpoint> {
Rect rect;
boolean isStart;
double val;
Endpoint(Rect r, boolean start, double v){
rect = r;
isStart = start;
val = v;
}
public int compareTo(Endpoint other){
int ret;
double diff= this.val - other.val;
if(diff == 0) ret = 0;
else ret = (diff>;0)?1:-1;
if(ret == 0){
ret = this.isStart?1:-1;
if(this.rect != other.rect){
ret = -ret;
}
}
return ret;
}
}
static class Rect {
Rect (double minX, double minY, double maxX, double maxY){
xStart = new Endpoint(this, true, minX);
xEnd = new Endpoint(this, false, maxX);
yStart = new Endpoint(this, true, minY);
yEnd = new Endpoint(this, false, maxY);
}
Endpoint xStart;
Endpoint xEnd;
Endpoint yStart;
Endpoint yEnd;
}
public static int getNumIntersecting(Rect [] rects){
TreeSet<Endpoint> sortedByX = new TreeSet<>();
for(Rect r: rects){
sortedByX.add(r.xStart);
sortedByX.add(r.xEnd);
}
TreeSet<Endpoint> sortedByY = new TreeSet<>();
int ret =0;
for(Endpoint ep: sortedByX){
Rect r = ep.rect;
if(ep.isStart){
ret+= countIntersections(sortedByY.subSet(r.yStart, false,r.yEnd, false),
r.yStart.val, r.yEnd.val);
sortedByY.add(r.yStart); sortedByY.add(r.yEnd);
} else {
sortedByY.remove(r.yStart); sortedByY.remove(r.yEnd);
}
}
return ret>;0?ret+1:0; // count the first square with whom an intersection is found
}
private static int countIntersections(Set<Endpoint> eps, double minY, double maxY){
HashSet<Rect> rects = new HashSet<>();
for(Endpoint ep : eps){
rects.add(ep.rect);
}
return rects.size();
}