leetcode/apps/docs/content/(03-hierarchical)/heap-priority-queue/973-k-closest-points-to-origin.mdx

---
title: '973. K Closest Points to Origin'
description: 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)
icon: dot
topics:
  - { name: "Array", slug: "array" }
  - { name: "Math", slug: "math" }
  - { name: "Divide and Conquer", slug: "divide-and-conquer" }
  - { name: "Geometry", slug: "geometry" }
  - { name: "Sorting", slug: "sorting" }
  - { name: "Heap (Priority Queue)", slug: "heap-priority-queue" }
  - { name: "Quickselect", slug: "quickselect" }
  - { name: "K-D Tree", slug: "k-d-tree" }
video: "rI2EBUEMfTk"
sidebar:
  label: 'K Closest Points to Origin'
  badge: 'Medium'
---

### 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 <= 10^4`
- `-10^4 <= xi, yi <= 10^4`

## Approach

```mermaid
flowchart TD
  S(["kClosest(points, k)"]) --> I["minHeap = []"]
  I --> F{"more x, y in points?"}
  F -- yes --> D["dist = x**2 + y**2 — squared distance orders the same as sqrt"]
  D --> A["minHeap.append([dist, x, y])"]
  A --> F
  F -- no --> H["heapq.heapify(minHeap), res = [] — O(n) build, smallest dist on top"]
  H --> W{"k > 0?"}
  W -- yes --> P["dist, x, y = heapq.heappop(minHeap)"]
  P --> R["res.append([x, y]), k -= 1"]
  R --> W
  W -- no --> E(["return res"])
```

## Solution

```py
import heapq


class Solution:
    def kClosest(self, points: List[List[int]], k: int) -> List[List[int]]:
        minHeap = []
        for x, y in points:
            dist = (x**2) + (y**2)
            minHeap.append([dist, x, y])

        heapq.heapify(minHeap)
        res = []

        while k > 0:
            dist, x, y = heapq.heappop(minHeap)
            res.append([x, y])
            k -= 1

        return res
```