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K_closest_points_to_origin.py
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'''
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.
Constraints:
1 <= k <= points.length <= 104
-104 < xi, yi < 104
'''
import math
class Solution:
def kClosest(self, points: List[List[int]], k: int) -> List[List[int]]:
distance_arr=[]
#calculating the distance from center
for i in points:
distance_arr.append(math.sqrt(pow(i[0],2)+pow(i[1],2)))
#Mapping the values of distance and point
#Also considering the collision if two points have same distance from origin
hash_map={}
for i in range(len(points)):
if(distance_arr[i] in hash_map):
hash_map[distance_arr[i]].append(points[i])
else:
hash_map[distance_arr[i]] = [points[i]]
#sorting distances
distance_arr.sort()
#Retrieving the values in the order of distance from distance_arr
final_arr,count = [],0
for i in range(len(points)):
for j in hash_map[distance_arr[i]]:
final_arr.append(j)
count+=1
if(count>=k):
break
return final_arr