K Closest Points to Origin

Problem

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:

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]]. 

Solution Approach

Expected Time Complexity: $O(nlogn)$

Click - to see solution code

C++

class Solution {
   public:
    vector<vector<int>> kClosest(vector<vector<int>>& points, int k) {
        int n = points.size();
        vector<vector<double>> dis(n, vector<double>(2));
        vector<vector<int>> ans(k, vector<int>(2));

        for (int j = 0; j < n; j++) {
            double a =
                (points[j][0] * points[j][0] + points[j][1] * points[j][1]);
            a = sqrt(a);
            dis[j][0] = a;
            dis[j][1] = j;
        }
        sort(dis.begin(), dis.end());
        for (int i = 0; i < k; i++) {
            ans[i][0] = points[dis[i][1]][0];
            ans[i][1] = points[dis[i][1]][1];
        }

        return ans;
    }
};