[Programming Problem] K Closest Points to Origin

Example 1:
Input: points = [[1,3],[-2,2]], k = 1
Output: [[-2,2]]
Explanation:
The distance between (1, 3) and the origin is sqrt(10).
The distance between (-2, 2) and the origin is sqrt(8).
Since sqrt(8) < sqrt(10), (-2, 2) is closer to the origin.
We only want the closest k = 1 points from the origin, so the answer is just [[-2,2]].
 
Example 2:
Input: points = [[3,3],[5,-1],[-2,4]], k = 2
Output: [[3,3],[-2,4]]
Explanation: The answer [[-2,4],[3,3]] would also be accepted.

Example 1: Input: points = [[1,3],[-2,2]], k = 1 Output: [[-2,2]] Explanation: The distance between (1, 3) and the origin is sqrt(10). The distance between (-2, 2) and the origin is sqrt(8). Since sqrt(8) < sqrt(10), (-2, 2) is closer to the origin. We only want the closest k = 1 points from the origin, so the answer is just [[-2,2]]. Example 2: Input: points = [[3,3],[5,-1],[-2,4]], k = 2 Output: [[3,3],[-2,4]] Explanation: The answer [[-2,4],[3,3]] would also be accepted.

import java.util.*;
 
class Point {
    int x;
    int y;
    double dist;
    public Point(int x, int y) {
        this.x = x;
        this.y = y;
        this.dist = Math.sqrt(Math.pow(x * 1.0, 2) + Math.pow(y * 1.0, 2));
    }
}
 
class Solution {
    public int[][] kClosest(int[][] points, int k) {
        PriorityQueue maxHeap = new PriorityQueue<Point>((x, y) -> Double.compare(y.dist, x.dist));
        for (int i = 0 ; i < points.length ; i++ ) {
            maxHeap.offer(new Point(points[i][0], points[i][1]));
            if (maxHeap.size() >= k+1) maxHeap.poll();
        }
 
        int ret[][] = new int[maxHeap.size()][2];
        int i = 0;
        while (maxHeap.size() > 0) {
            Point p = (Point) maxHeap.poll();
            ret[i][0] = p.x;
            ret[i][1] = p.y;
            i++;
        }
 
        return ret;
    }
}

import java.util.*; class Point { int x; int y; double dist; public Point(int x, int y) { this.x = x; this.y = y; this.dist = Math.sqrt(Math.pow(x * 1.0, 2) + Math.pow(y * 1.0, 2)); } } class Solution { public int[][] kClosest(int[][] points, int k) { PriorityQueue maxHeap = new PriorityQueue<Point>((x, y) -> Double.compare(y.dist, x.dist)); for (int i = 0 ; i < points.length ; i++ ) { maxHeap.offer(new Point(points[i][0], points[i][1])); if (maxHeap.size() >= k+1) maxHeap.poll(); } int ret[][] = new int[maxHeap.size()][2]; int i = 0; while (maxHeap.size() > 0) { Point p = (Point) maxHeap.poll(); ret[i][0] = p.x; ret[i][1] = p.y; i++; } return ret; } }