# https://leetcode.com/problems/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:
# 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.

class Solution:
    def kClosest(self, points: List[List[int]], k: int) -> List[List[int]]:
        def getdistance(x, y):
            import math
            return math.sqrt(x * x + y * y)

        d = {}

        for i in points:
            newdistance = getdistance(*i)
            if newdistance in d:
                d[newdistance].append(i)
            else:
                d[newdistance] = [i]

        keysorted = sorted(d)
        lst = []
        for i in keysorted:
            if k > 0:
                if len(d[i]) > k:
                    lst.extend(d[i][:k])
                    break
                elif len(d[i]) == k:
                    lst.extend(d[i])
                    break
                else:
                    lst.extend(d[i])
                    k -= len(d[i])
        return lst

#get one from someone
class Solution:
    def kClosest(self, points: List[List[int]], k: int) -> List[List[int]]:
        return sorted(points, key = lambda x: sqrt(x[0]**2 + x[1]**2))[0:k]