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# K Closest Points to Origin LeetCode Solution

## Problem – K Closest Points to Origin

Given an array of `points` where `points[i] = [xi, yi]` 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., `√(x1 - x2)2 + (y1 - y2)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 <= 104`
• `-104 < xi, yi < 104`

### K Closest Points to Origin LeetCode Solution in Java

``````    public int[][] kClosest(int[][] points, int K) {
Arrays.sort(points, Comparator.comparing(p -> p * p + p * p));
return Arrays.copyOfRange(points, 0, K);
}``````

### 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 * a + a * a < b * b + b * b;
});
return vector<vector<int>>(A.begin(), A.begin() + K);
}``````
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