# leetcode\_973

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 <= K <= points.length <= 10000 -10000 < points\[i]\[0] < 10000 -10000 < points\[i]\[1] < 10000

## Solutions

* Select top-k problem
* **max heap**

```cpp
class Solution {
public:
    vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
        if (points.size() <= K)
            return points;
        auto cmp = [&](int i1, int i2) {
            auto & p1 = points[i1];
            auto & p2 = points[i2];
            return p1[0] * p1[0] + p1[1] * p1[1] 
                 < p2[0] * p2[0] + p2[1] * p2[1];
        };
        priority_queue<int, vector<int>, decltype(cmp)> pq(cmp);

        for (int i = 0; i < K; i++)
            pq.push(i);

        for (int i = K; i < points.size(); i++) {
            pq.push(i);
            pq.pop();
        }

        vector<vector<int>> res;
        while (pq.size()) {
            res.push_back(points[pq.top()]);
            pq.pop();
        }

        return res;
    }
};
```

1. **quick select**
2. Use stl's nth\_element as example

```cpp
class Solution {
public:
    vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
        if (points.size() <= K)
            return points;
        auto cmp = [&](auto & p1, auto & p2) {
            return p1[0] * p1[0] + p1[1] * p1[1] 
                 < p2[0] * p2[0] + p2[1] * p2[1];
        };
        nth_element(
            points.begin(), 
            points.begin() + K, 
            points.end(), 
            cmp
        );

        return {points.begin(), points.begin() + K};
    }
};
```