10. K Closest Points to Origin

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).

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.

Solution: (Heap)

class Solution
{
public:
    vector<vector<int>> kClosest(vector<vector<int>> &points, int k)
    {

        vector<vector<int>> res;
        priority_queue<pair<double, int>, vector<pair<double, int>>, greater<pair<double, int>>> pq;

        for (int i = 0; i < points.size(); i++)
        {
            int x = points[i][0];
            int y = points[i][1];

            double dist = sqrt(x * x + y * y);
            
            pq.push({dist, i});
        }

        while (k)
        {
            int idx = pq.top().second;
            res.push_back(points[idx]);
            k--;
            pq.pop();
        }

        return res;
    }
};