2021-12-26 Daily-Challenge

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Today I have done leetcode's December LeetCoding Challenge with cpp.

December LeetCoding Challenge 26

Description

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).

Example 1:

img

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 <= 104
  • -104 < xi, yi < 104

Solution

auto speedup = []() {
  std::ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  return nullptr;
}();
class Solution {
public:
  vector<vector<int>> kClosest(vector<vector<int>>& points, int k) {
    nth_element(points.begin(), points.begin() + k - 1, points.end(), [&](const vector<int> &a, const vector<int> &b) {
      return a[0] * a[0] + a[1] * a[1] <
             b[0] * b[0] + b[1] * b[1];
    });
    points.erase(points.cbegin() + k, points.cend());
    return points;
  }
};

// Accepted
// 85/85 cases passed (92 ms)
// Your runtime beats 100 % of cpp submissions
// Your memory usage beats 98.48 % of cpp submissions (49.1 MB)
Algorithm