Given an array of points where points[i] = [x_{i}, y_{i}] 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., √(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}).

You may return the answer in any order. The answer is guaranteed to be unique (except for the order that it is in).

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 <= 10^{4}

-10^{4} < x_{i}, y_{i} < 10^{4}

K Closest Points to Origin LeetCode Solution in Java

K Closest Points to Origin LeetCode Solution in Python

def kClosest(self, points, K):
return heapq.nsmallest(K, points, lambda (x, y): x * x + y * y)

K Closest Points to Origin LeetCode Solution in C++

vector<vector<int>> kClosest(vector<vector<int>>& A, int K) {
nth_element(A.begin(), A.begin() + K, A.end(), [](vector<int>& a, vector<int>& b) {
return a[0] * a[0] + a[1] * a[1] < b[0] * b[0] + b[1] * b[1];
});
return vector<vector<int>>(A.begin(), A.begin() + K);
}

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