K Closest Points to Origin

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2020-05-30

973. K Closest Points to Origin

Question:

We have a list of points on the plane. Find the K closest points to the origin (0, 0).

(Here, the distance between two points on a plane is the Euclidean distance.)

You may return the answer in any order. The answer is guaranteed to be unique (except for the order that it is in.)

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]]
(The answer [[-2,4],[3,3]] would also be accepted.)

Note:

1 <= K <= points.length <= 10000
-10000 < points[i][0] < 10000
-10000 < points[i][1] < 10000

Solution:

Using Priority Queue with distance in its comparator.

Time: O((N*lgK)
Space: O(N) We can achieve to O(K) by checking the size when insert.

class Solution {
    public int[][] kClosest(int[][] points, int K) {
        int[][] result = new int[K][2];
        if (points == null || points.length == 0 || points.length < K) return result;
        PriorityQueue<Point> pq = new PriorityQueue<>(K, new Comparator<Point>(){
            @Override
            public int compare(Point a, Point b) {
                int a_dis = a.x * a.x + a.y * a.y;
                int b_dis = b.x * b.x + b.y * b.y;
                return a_dis - b_dis;
            }
        });
        
        for (int[] i : points) {
            pq.add(new Point(i[0], i[1]));
        }
        
        for (int i = 0; i < K; i++) {
            Point curr = pq.poll();
            result[i][0] = curr.x;
            result[i][1] = curr.y;
        }
        
        return result;
    }
}

class Point {
    int x;
    int y;
    public Point(int xaxis, int yaxis){
        this.x = xaxis;
        this.y = yaxis;
    }
}