973.k-closest-points-to-origin.cpp

//  Tag: Array, Math, Divide and Conquer, Geometry, Sorting, Heap (Priority Queue), Quickselect
//  Time: O(NlogN)
//  Space: O(K)
//  Ref: -
//  Note: -

//  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.
//  
//   
//  Constraints:
//  
//  1 <= k <= points.length <= 104
//  -104 <= xi, yi <= 104
//  
//  

class Solution {
public:
    vector<vector<int>> kClosest(vector<vector<int>>& points, int k) {
        priority_queue<pair<int, int>> heap;
        for (int i = 0; i < points.size(); i++) {
            int distance = points[i][0] * points[i][0] + points[i][1] * points[i][1];
            if (heap.size() == k && distance < heap.top().first) {
                heap.pop();
            }

            if (heap.size() < k) {
                heap.emplace(distance, i);
            }
        }

        vector<vector<int>> res;
        while (!heap.empty()) {
            vector<int> p = points[heap.top().second];
            heap.pop();
            res.push_back(p);
        }

        return res;
    }
};