## Problem Statement#

### 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)`.

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).

### Examples#

#### 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`

## Key Insights#

This is a pretty straight forward problem. It’s listed as a medium on Leetcode, but I think this one is probably more of an easy if you know how to perform sorting in Python or if you know how to use the `heapq` library.

The easiest approach is to use `*.sort(...)` to order the list by distance with a lambda function. Alternatively, the coordinates and their associated distances could be added to a min heap as tuples and the `k` closest coordinates could be popped from the heap and returned.

## Python Solution#

### Sorting Solution#

``````class Solution:
def kClosest(self, points: List[List[int]], k: int) -> List[List[int]]:
"""
T: O(N log N )
S: O(1)

- The euclidean distance requires a square root.
However, since we only care about the sort order,
we can save an operation by not taking the square
root of the result.
- Notice how we can use the key = lambda x : ...
to create a custom sort function
- Once we have sorted the list, we can return the
k - closest points to the origin

"""
points.sort(key = lambda x: x**2 + x**2)
return points[:k]
``````

### Heap Solution#

``````from heapq import heappush, heappop, heapify

class Solution:
def kClosest(self, points: List[List[int]], k: int) -> List[List[int]]:
"""
T: O( log N)
S: O( N )

- This is not the most efficient solution, but an interviewer
could ask you use a heap. Using *.sort(...) will execute faster
due to the underlying implementation
- Loop over the points list, calculate the distance, and
push a distance, coordinate tuple onto the min heap
- Pop the first k elements off of the heap and return them
as a list.
"""
distances = []
heapify(distances)

for x, y in points:
dist = (x - 0)**2 + (y - 0)**2
heappush(distances,  (dist, [x, y]))

out = []

for i in range(k):
_, point = heappop(distances)
out.append(point)

return out
``````