1014. K Closest Points to Origin

• Difficulty: Medium

• Topics: Divide and Conquer, Heap, Sort

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Problem:

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][0] < 10000
3. -10000 < points[i][1] < 10000

Solutions:

class Solution {
public:
    vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
        auto comparator = [](vector<int> p1, vector<int> p2) {
            return p1[0] * p1[0] + p1[1] * p1[1] < p2[0] * p2[0] + p2[1] * p2[1];
        };
        priority_queue<vector<int>, vector<vector<int>>, decltype(comparator)> pq(comparator);

        for (auto& point : points) {
            pq.push(point);
            if (pq.size() > K) {
                pq.pop();
            }
        }

        vector<vector<int>> ret;
        while (!pq.empty()) {
            ret.push_back(pq.top()); pq.pop();
        }

        return ret;
    }
};