# K Closest Points To Origin Problem

## Description

LeetCode Problem 973.

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:

``````1
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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:

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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

## Sample C++ Code

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class Solution {
public:
vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
int n = points.size();
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>> > pq;

for (int i = 0; i < n; i ++) {
pq.push({points[i][0] * points[i][0] + points[i][1] * points[i][1], i});
}

vector<vector<int>> ans;
while (K > 0) {
ans.push_back(points[pq.top().second]);
pq.pop();
K --;
}
return ans;
}
};
``````