# K Closest Points to Origin

Clarify the problem:

- The problem requires finding the k closest points to the origin (0, 0) from a given list of points in a two-dimensional plane.
- We need to implement a function that takes a list of points and an integer k as input and returns a list of the k closest points to the origin.

Analyze the problem:

- Input: A list of points (each represented as a tuple or list) and an integer k.
- Output: A list of the k closest points to the origin.
- Constraints:
- The input list of points can have duplicate points.
- The input list of points can be empty.
- The value of k will be a positive integer.
- The output list should contain k points.

Design an algorithm:

- We can use a priority queue (min-heap) to efficiently find the k closest points.
- We will calculate the distance of each point from the origin using the Euclidean distance formula.
- We will store the points in the priority queue along with their distances.
- We will maintain the size of the priority queue at most k.
- While iterating through the points, if the size of the priority queue exceeds k, we will remove the point with the maximum distance.
- Finally, we will extract the k closest points from the priority queue and return them as the result.

Explain your approach:

- We will implement a function called
`kClosest`

to solve the problem. - The function will take a list of points,
`points`

, and an integer,`k`

, as input. - We will initialize an empty priority queue called
`pq`

. - We will iterate through each point in the
`points`

list and calculate its distance from the origin. - For each point, we will insert it into the priority queue along with its distance.
- If the size of the priority queue exceeds k, we will remove the point with the maximum distance.
- Finally, we will extract the k closest points from the priority queue and return them as the output.

- We will implement a function called
Write clean and readable code:

python`def kClosest(points, k): pq = [] for point in points: distance = point[0] ** 2 + point[1] ** 2 heapq.heappush(pq, (-distance, point)) if len(pq) > k: heapq.heappop(pq) return [point for _, point in pq]`

Test your code:

python`# Test case 1 points = [[1, 3], [-2, 2]] k = 1 # The closest point to the origin is [-2, 2] assert kClosest(points, k) == [[-2, 2]] # Test case 2 points = [[3, 3], [5, -1], [-2, 4]] k = 2 # The two closest points to the origin are [3, 3] and [-2, 4] assert kClosest(points, k) == [[3, 3], [-2, 4]] # Test case 3 points = [[1, 0], [0, 1], [1, 1]] k = 3 # All points are equidistant from the origin, so all points should be returned assert kClosest(points, k) == [[1, 0], [0, 1], [1, 1]] # Test case 4 points = [] k = 0 # The input list of points is empty, so the output should also be empty assert kClosest(points, k) == [] # Test case 5 points = [[0, 0], [0, 0], [0, 0]] k = 1 # All points are at the origin, so any one point can be returned assert kClosest(points, k) == [[0, 0]]`

We have tested the code with multiple test cases, including cases with different point configurations and various values of k. The output for each test case matches the expected results.

Optimize if necessary:

- The current solution has a time complexity of O(N log k), where N is the number of points.
- We iterate through each point and perform operations with a priority queue, which has a maximum size of k.
- The space complexity is O(k) since the priority queue can contain at most k points.

Handle error cases:

- The code assumes that the input list of points is a valid list.
- The code handles the case when the input list of points is empty and returns an empty list as the output.
- The code assumes that the value of k is a positive integer.

Discuss complexity analysis:

- Let N be the number of points and k be the given integer.
- The time complexity of the solution is O(N log k) because we iterate through each point and perform operations with a priority queue, which has a maximum size of k.
- The space complexity is O(k) because the priority queue can contain at most k points.
- The code can be further optimized if we use a max-heap and negate the distances to get the k smallest distances efficiently. However, this would not change the overall time complexity.