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Priority_queue - Question 4

973 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).


Constraints:

1 <= k <= points.length <= 10^4

-10^4 < xi, yi < 10^4


Analysis:

It is quite straightforward to find the K points that have the closest distances to the original point.

Since we need to find the smallest values, the root of the priority_queue (pq) is the largest one for all the elements inside the pq.

Once the size of pq is larger than K, we need to pop out one.


See the code below:

class Solution {
public:
    typedef pair<int, int> pp;
    vector<vector<int>> kClosest(vector<vector<int>>& points, int k) {
        vector<vector<int>> res;
        int n = points.size();
        priority_queue<pp> pq;
        for(int i=0; i<n; ++i) {
            int x = points[i][0], y = points[i][1];
            pq.push({x*x + y*y, i});
            if(pq.size() > k) pq.pop();
        }
        while(pq.size()) {
            int id = pq.top().second;
            res.push_back(points[id]);
            pq.pop();
        }
        return res;
    }
};



Upper Layer

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