973. K Closest Points to Origin

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973. K Closest Points to Origin

Description

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:

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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 <= xi, yi <= 10^4

Hints/Notes

Solution

Language: C++

quick select

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class Solution {
public:
    vector<vector<int>> kClosest(vector<vector<int>>& points, int k) {
        quick_select(points, k, 0, points.size() - 1);
        return vector<vector<int>>(points.begin(), points.begin() + k);
    }

    void quick_select(vector<vector<int>>& points, int k, int left, int right) {
        int pivot = right, pivotDist = points[pivot][0] * points[pivot][0] + points[pivot][1] * points[pivot][1];
        int i = left - 1;
        for (int j = left; j < right; j++) {
            int dist = points[j][0] * points[j][0] + points[j][1] * points[j][1];
            if (dist <= pivotDist) {
                swap(points[++i], points[j]);
            }
        }
        i++;
        swap(points[i], points[right]);
        // from left to i, a total of i - left + 1 numbers are <= pivot
        if (i - left + 1 > k) {
            quick_select(points, k, left, i - 1);
        } else if (i - left + 1 < k) {
            quick_select(points, k - i + left - 1, i + 1, right);
        }
    }
};

priority queue:

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