K Closest Points to Origin
Problem Statement - link #
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).
Examples #
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 < xi, yi < 10^4
Solutions #
class Solution {
public:
vector<vector<int>> kClosest(vector<vector<int>>& points, int k) {
// Sort the vector with a custom lambda comparator function
sort(points.begin(), points.end(), [&](vector<int>& a, vector<int>& b) {
return squaredDistance(a) < squaredDistance(b);
});
// Return the first k elements of the sorted vector
return vector<vector<int>>(points.begin(), points.begin() + k);
}
private:
int squaredDistance(vector<int>& point) {
// Calculate and return the squared Euclidean distance
return point[0] * point[0] + point[1] * point[1];
}
};