Top K Frequent Closest Points

You are given an array of two-dimensional integer points. Equal coordinate pairs represent repeated observations of the same distinct point.
Return the k distinct points with the highest frequencies, ordered by:
higher frequency first;
smaller squared Euclidean distance x*x + y*y from the origin;
smaller x coordinate;
smaller y coordinate.
Return each selected point once.

Function
topKFrequentClosest(points: int[][], k: int) → int[][]

Examples
Example 1
points = [[1,1],[1,1],[2,0],[2,0],[2,0],[0,3]]
k = 2
return = [[2,0],[1,1]]
[2,0] occurs three times and [1,1] occurs twice, so frequency fixes their order.
Example 2
points = [[3,0],[0,2],[-2,0],[3,0],[0,2],[-2,0]]
k = 3
return = [[-2,0],[0,2],[3,0]]
All three points occur twice. The two distance-four points tie on distance, so smaller x puts [-2,0] first; [3,0] is farther away.

Constraints
1 <= points.length <= 200000
Every row has exactly two integers x and y with -100000 <= x, y <= 100000.
1 <= k <= the number of distinct coordinate pairs.
Use signed 64-bit arithmetic for squared distances.

Step one: count occurrences of each point in a hash map keyed by the (x, y) pair. That gives you distinct points with frequencies. Step two: order them by the tuple (-frequency, x*x + y*y, x, y). You can sort all distinct points with that key, which is O(n log n) and perfectly fine for 200000. Or keep a heap of size k with the inverse comparator for O(n log k). Sorting is the lower-risk choice. The pitfalls are all small. Squared distance can reach 2 * 10^10, so use 64-bit. Don't mix up the direction of each tie-break. Return each point once, not once per occurrence. If you freeze on the comparator during the live OA, StealthCoder can give you the working solution on screen while you stay calm and keep typing.

The honest play: practice the pattern, and have StealthCoder ready for the one you didn't see coming.