Return the K Nearest Input Points to the Origin

### Problem Return the k points from `points` with smallest Euclidean distance to `(0,0)`. The selected collection is guaranteed unique; its returned order may be arbitrary. Compare distances using `sqrt(x*x+y*y)`, or equivalently their squared values. ```cpp class Solution { public: vector<vector<int>> kClosest(vector<vector<int>>& points, int k); }; ``` ### Examples **Example 1:** ```text Input: points = [[0,2],[3,0],[1,1]], k = 2 Output: [[1,1],[0,2]] Explanation: Squared distances are4,9,and2, so the points with distances squared2 and4 are selected. Their order may be reversed. ``` **Example 2:** ```text Input: points = [[4,0],[0,1]], k = 1 Output: [[0,1]] Explanation: Squared distance1 is smaller than16. ``` ### Constraints - `1 <= k <= points.length <= 10^4`. - `-10^4 <= x_i,y_i <= 10^4`. - The k-point answer is guaranteed unique except for its order. ### Contract considerations The answer is unique as a collection, not as an ordered array. The original any-order allowance remains intact. ```hint Read the boundary condition The square root is monotone, so squared distances give the same ranking. ``` ### What a Strong Answer Covers The answer is unique as a collection, not as an ordered array. The original any-order allowance remains intact. ### Follow-up Questions Optional practice extension, separate from the original task: How would you answer successive queries with different values of k on the same points?