找到最小半径，使得至少 k 点位于圆内

给定一个正整数 K，一个圆心在 (0, 0) 和一些点的坐标。任务是找到圆的最小半径，使至少 k 个点位于圆内。输出最小半径的平方。
例子：

Input : (1, 1), (-1, -1), (1, -1), 
         k = 3
Output : 2
We need a circle of radius at least 2
to include 3 points.


Input : (1, 1), (0, 1), (1, -1), 
         k = 2
Output : 1
We need a circle of radius at least 1
to include 2 points. The circle around
(0, 0) of radius 1 would include (1, 1)
and (0, 1).

这个想法是从原点 (0, 0) 找到每个点的欧几里得距离的平方。现在，按递增顺序对这些距离进行排序。现在距离的第 k^个元素是所需的最小半径。
下面是这个方法的实现：

C++

// C++ program to find minimum radius
// such that atleast k point lie inside
// the circle
#include
using namespace std;
 
// Return minumum distance required so that
// aleast k point lie inside the circle.
int minRadius(int k, int x[], int y[], int n)
{
   int dis[n];
     
   // Finding distance between of each
   // point from origin
   for (int i = 0; i < n; i++)
       dis[i] = x[i] * x[i] + y[i] * y[i];
     
    // Sorting the distance
    sort(dis, dis + n);
     
    return dis[k - 1];
}
 
// Driven Program
int main()
{
  int k = 3;
  int x[] = { 1, -1, 1 };
  int y[] = { 1, -1, -1 };
  int n = sizeof(x)/sizeof(x[0]);
     
  cout << minRadius(k, x, y, n) << endl;
     
  return 0;
}

Java

// Java program to find minimum radius
// such that atleast k point lie inside
// the circle
import java.util.Arrays;
 
class GFG
{
 
    // Return minumum distance required so that
    // aleast k point lie inside the circle.
    static int minRadius(int k, int[] x, int[] y,
                                          int n)
    {
        int[] dis=new int[n];
     
        // Finding distance between of each
        // point from origin
        for (int i = 0; i < n; i++)
            dis[i] = x[i] * x[i] + y[i] * y[i];
     
        // Sorting the distance
        Arrays.sort(dis);
     
        return dis[k - 1];
    }
 
    // Driven Program
    public static void main (String[] args) {
         
    int k = 3;
    int[] x = { 1, -1, 1 };
    int[] y = { 1, -1, -1 };
    int n = x.length;
     
    System.out.println(minRadius(k, x, y, n));
 
    }
}
 
/* This code is contributed by Mr. Somesh Awasthi */

Python3

# Python3 program to find minimum radius
# such that atleast k point lie inside
# the circle
 
 
# Return minumum distance required so
# that aleast k point lie inside the
# circle.
def minRadius(k, x, y, n):
    dis = [0] * n
 
    # Finding distance between of each
    # point from origin
 
    for i in range(0, n):
        dis[i] = x[i] * x[i] + y[i] * y[i]
 
    # Sorting the distance
    dis.sort()
 
    return dis[k - 1]
         
# Driver Program
k = 3
x = [1, -1, 1]
y = [1, -1, -1]
n = len(x)
 
print(minRadius(k, x, y, n))
 
# This code is contributed by
# Prasad Kshirsagar

C#

// C# program to find minimum radius
// such that atleast k point lie inside
// the circle
using System;
 
class GFG {
 
    // Return minumum distance required
    // so that aleast k point lie inside
    // the circle.
    static int minRadius(int k, int []x,
                          int[] y, int n)
    {
        int[] dis = new int[n];
     
        // Finding distance between of
        // each point from origin
        for (int i = 0; i < n; i++)
            dis[i] = x[i] * x[i] +
                       y[i] * y[i];
     
        // Sorting the distance
        Array.Sort(dis);
     
        return dis[k - 1];
    }
 
    // Driven Program
    public static void Main ()
    {
        int k = 3;
        int[] x = { 1, -1, 1 };
        int[] y = { 1, -1, -1 };
        int n = x.Length;
         
        Console.WriteLine(
              minRadius(k, x, y, n));
    }
}
 
// This code is contributed by vt_m.

Javascript

// Javascript program to find minimum radius
// such that atleast k point lie inside
// the circle
 
    // Return minumum distance required so that
    // aleast k polet lie inside the circle.
    function minRadius(k, x, y, n) {
     
        let dis = Array.from({length: n}, (_, i) => 0);
       
        // Finding distance between of each
        // polet from origin
        for (let i = 0; i < n; i++)
            dis[i] = x[i] * x[i] + y[i] * y[i];
       
        // Sorting the distance
        dis.sort();
       
        return dis[k - 1];
    }
   
// driver function
 
    let k = 3;
    let x = [ 1, -1, 1 ];
    let y = [ 1, -1, -1 ];
    let n = x.length;
       
    document.write(minRadius(k, x, y, n));
  
 // This code is contributed by code_hunt.

输出：

2

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