We use a Max-Heap of size k. We store the squared distance (x^2 + y^2) and the point itself. By using a Max-Heap, we can always remove the point that is furthest from the origin among the current k points.

1. Initialize a max-heap.
2. For each (x, y) in points:
   • Calculate dist = -(x*x + y*y) (negative to simulate Max-Heap in Python).
   • Push (dist, [x, y]) to the heap.
   • If heap size > k, pop the largest distance (the top).
3. Extract points from the heap and return.

• Time Complexity: O(N log K).
• Space Complexity: O(K).

import heapq

def k_closest(points, k):
    max_heap = []
    
    for x, y in points:
        dist = x*x + y*y
        # Store negative distance for max-heap behavior
        heapq.heappush(max_heap, (-dist, [x, y]))
        
        if len(max_heap) > k:
            heapq.heappop(max_heap)
            
    return [p[1] for p in max_heap]