# LeetCode 973. 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:**![](https://assets.leetcode.com/uploads/2021/03/03/closestplane1.jpg)

```
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`

## Solution:

[English Version in Youtube](https://youtu.be/V2pdqKe2mjY)

[中文版解答Youtube Link](https://youtu.be/HaRE_RS70IQ)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV1s54y1a7iQ/)

```
class Solution {
public:
    vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
        auto cmp = [&](const vector<int>& a, const vector<int>& b) {
            return a[0] * a[0] + a[1] * a[1] <  b[0] * b[0] + b[1] * b[1]; 
        };
        
        nth_element(points.begin(), points.begin() + K, points.end(), cmp);
        // sort(points.begin(), points.end(), cmp);
        
        vector<vector<int>> ans(points.begin(), points.begin() + K);
        return ans;
    }
};
```


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