Problem Description
Given an array of points where points[i] = [xi, yi] represents a point on the X-Y plane and an integer k, return the k closest points to the origin (0, 0).
The distance between two points on the X-Y plane is the Euclidean distance (i.e., √(x1 - x2)2 + (y1 - y2)2).
You may return the answer in any order. The answer is guaranteed to be unique (except for the order that it is in).
Solution
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class Solution {
public double getDistanceFromOrigin(double x, double y){
return Math.sqrt(x*x + y*y);
}
public int[][] kClosest(int[][] points, int k) {
PriorityQueue<Pair> pq = new PriorityQueue<>();
for(int i=0; i<points.length; i++){
int x = points[i][0];
int y = points[i][1];
double dist = getDistanceFromOrigin(x, y);
pq.add(new Pair(dist, new int[]{x,y}));
}
int[][] ans = new int[k][2];
for(int i=0; i<k; i++){
Pair p = pq.poll();
int x = p.coordinates[0];
int y = p.coordinates[1];
ans[i] = new int[]{x,y};
}
return ans;
}
}
class Pair implements Comparable<Pair>{
double distance;
int[] coordinates;
Pair(double d, int[] c){
this.distance = d;
this.coordinates = c;
}
public int compareTo(Pair p){
if(this.distance < p.distance) return -1;
if(this.distance > p.distance) return 1;
return 0;
}
}
Another way to code
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class Solution {
public int distance(int x, int y){
return x*x + y*y; //we don't necessarily need square root here since all we need to do is sort. Math.sqrt() will also work.
}
public int[][] kClosest(int[][] points, int k) {
//Compare two arrays based on their first element at 0th index.
PriorityQueue<Integer[]> pq = new PriorityQueue<>((a,b) -> a[0]-b[0]);
for(int i=0; i<points.length; i++){
Integer x = points[i][0];
Integer y = points[i][1];
pq.add(new Integer[]{distance(x,y), x, y});
}
int[][] ans = new int[k][2];
for(int i=0; i<k; i++){
Integer[] p = pq.poll();
ans[i] = new int[]{p[1], p[2]};
}
return ans;
}
}