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# 973. K Closest Points to Origin

## 973. [K Closest Points to Origin](https://leetcode.com/problems/k-closest-points-to-origin/)

## 1. Question

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)

<pre><code><strong>Input: points = [[1,3],[-2,2]], k = 1
</strong><strong>Output: [[-2,2]]
</strong><strong>Explanation:
</strong>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) &#x3C; 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]].
</code></pre>

**Example 2:**

<pre><code><strong>Input: points = [[3,3],[5,-1],[-2,4]], k = 2
</strong><strong>Output: [[3,3],[-2,4]]
</strong><strong>Explanation: The answer [[-2,4],[3,3]] would also be accepted.
</strong></code></pre>

&#x20;

**Constraints:**

* `1 <= k <= points.length <= 104`
* `-104 <= xi, yi <= 104`

## 2. Implementation

**(1) Heap**

```python
class Solution:
    def kClosest(self, points: List[List[int]], k: int) -> List[List[int]]:
        pq = []
        for x, y in points:
            dist = -(x*x + y*y)

            if len(pq) == k:
                heapq.heappushpop(pq, (dist, x, y))
            else:
                heapq.heappush(pq, (dist, x, y))
        return [(x, y) for (dist, x, y) in pq]
```

## 3. Time & Space Complexity

**Heap:** 时间复杂度O(n \* log(k)), 空间复杂度O(k)