Amazon

Apple

Asana

Oracle

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. `1 <= K <= points.length <= 10000`
2. `-10000 < points[i] < 10000`
3. `-10000 < points[i] < 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];
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) {
}

for (int i = 0; i < K; i++) {
Point curr = pq.poll();
result[i] = curr.x;
result[i] = curr.y;
}

return result;
}
}

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