Day 22 of 100 days dsa coding challenge

Taking on a new challenge: solving GeeksforGeeks POTD daily and sharing my solutions! 💻🔥
The goal: sharpen problem-solving skills, level up coding, and learn something new every day. Follow my journey! 🚀

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Problem:

https://www.geeksforgeeks.org/problems/k-closest-points-to-origin--172242/1

K Closest Points to Origin

Difficulty: Medium Accuracy: 62.4%

Given an integer k and an array of points points[][], where each point is represented as points[i] = [xi, yi] on the X-Y plane. Return the k closest points to the origin (0, 0).
The distance between two points on the X-Y plane is the Euclidean distance, defined as:
distance = sqrt( (x2 - x1)2 + (y2 - y1)2 )
Note: You can return the k closest points in any order, the driver code will print them in sorted order.
Test Cases are generated such that there will be a unique ans.

Examples:
Input: k = 2, points[] = [[1, 3], [-2, 2], [5, 8], [0, 1]]
Output: [[-2, 2], [0, 1]]
Explanation: The Euclidean distances from the origin are:
Point (1, 3) = sqrt(10)
Point (-2, 2) = sqrt(8)
Point (5, 8) = sqrt(89)
Point (0, 1) = sqrt(1)
The two closest points to the origin are [-2, 2] and [0, 1].
Input: k = 1, points = [[2, 4], [-1, -1], [0, 0]]
Output: [[0, 0]]
Explanation: The Euclidean distances from the origin are:
Point (2, 4) = sqrt(20)
Point (-1, -1) = sqrt(2)
Point (0, 0) = sqrt(0)
The closest point to origin is (0, 0).
Constraints:
1 ≤ k ≤ points.size() ≤ 105
-3*104 ≤ xi, yi ≤ 3*104

Solution:
import heapq

class Solution:
def kClosest(self, points, k):
heap = []
for x, y in points:
dist = x*x + y*y
heapq.heappush(heap, (dist, [x, y]))
res = []
for _ in range(k):
res.append(heapq.heappop(heap)[1])
return res