Detect Squares
You are given a 2D array of points where each point is represented as she,Question and said integers. Each point can be the vertex of a square with a side length of 0 or greater. Write a function that takes these points as input and returns the number of squares that can be formed using these points.
Constraints:
- 1 <= points.length <= 10^5
- -10^4 <= xi, yi <= 10^4
Examples:
Input: [[1,1],[2,2],[1,2],[2,1]]
Output: 1
Explanation: The points (1,1), (1,2), (2,1), and (2,2) form a square with a side length of 1.
Solutions
Hash Map
The provided solution uses a hash map to store the points and their frequencies. It then iterates over all pairs of points to calculate the distance between them. If the distance is not zero, it checks if the other two points that would form a square with the current pair are present in the hash map. If they are, it increments the count of squares.
public int countSquares(int[][] points) {
int count = 0;
Set<String> pointSet = new HashSet<>();
for (int[] point : points) {
pointSet.add(point[0] + "," + point[1]);
}
for (int i = 0;
i < points.length;
i++) {
for (int j = i + 1;
j < points.length;
j++) {
double distance = Math.sqrt(Math.pow(points[i][0] - points[j][0], 2) + Math.pow(points[i][1] - points[j][1], 2));
if (distance != 0) {
int[] p3 = {
points[i][0] + points[j][0] - points[i][0], points[i][1] + points[j][1] - points[i][1]}
;
int[] p4 = {
points[i][0] + points[j][0] - points[j][0], points[i][1] + points[j][1] - points[j][1]}
;
if (pointSet.contains(p3[0] + "," + p3[1]) && pointSet.contains(p4[0] + "," + p4[1])) {
count++;
}
}
}
}
return count;
}Follow-up:
How would you optimize the solution to handle a large number of points?
