找到恰好在给定正方形内的一个点

给定一个整数K和一个数组arr，每个元素x代表一个正方形，其两个顶点为(0，0)和(x，x) 。任务是找到一个恰好位于K个正方形中的点。
例子：

Input: arr[] = {1, 2, 3, 4}, K = 2 
Output: (3, 3) 
The point (3, 3) lies inside 3rd and 4th square only.
Input: arr[] = {8, 1, 55, 90}, K = 3 
Output: (8, 8) 
 

方法：由于所有正方形都有一个公共拐角点(0，0)，因此任何位于任何正方形中的点也将位于任何较大的正方形中。因此，我们可以简单地打印第K个最大正方形的另一个角。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
int PointInKSquares(int n, int a[], int k)
{
    sort(a, a + n);
    return a[n - k];
}
 
// Driver Program to test above function
int main()
{
    int k = 2;
    int a[] = { 1, 2, 3, 4 };
    int n = sizeof(a) / sizeof(a[0]);
 
    int x = PointInKSquares(n, a, k);
    cout << "(" << x << ", " << x << ")";
}

Java

// Java implementation of the approach
 
import java.io.*;
import java.util.*;
class GFG {
 
 
static int PointInKSquares(int n, int a[], int k)
{
    Arrays.sort(a);
    return a[n - k];
}
 
// Driver Program to test above function
 
    public static void main (String[] args) {
            int k = 2;
    int []a = { 1, 2, 3, 4 };
    int n = a.length;
 
    int x = PointInKSquares(n, a, k);
    System.out.println( "(" + x + ", " + x +")");
 
     
    }
}
// This code is contributed by anuj_67..

Python3

# Python 3 implementation of the
# above approach
def PointInKSquares(n, a, k) :
     
    a.sort()
    return a[n - k]
 
# Driver Code
if __name__ == "__main__" :
     
    k = 2
    a = [1, 2, 3, 4]
    n = len(a)
     
    x = PointInKSquares(n, a, k)
    print("(", x, ",", x, ")")
 
# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;
class GFG
{
 
static int PointInKSquares(int n,
                           int []a, int k)
{
    Array.Sort(a);
    return a[n - k];
}
 
// Driver Code
public static void Main (String[] args)
{
    int k = 2;
    int []a = { 1, 2, 3, 4 };
    int n = a.Length;
     
    int x = PointInKSquares(n, a, k);
    Console.WriteLine("(" + x + ", " + x +")");
}
}
 
// This code is contributed
// by Arnab Kundu

Javascript

 
// Javascript implementation of the approach
 
 
function PointInKSquares(n, a, k)
{
    a.sort();
    return a[n - k];
}
 
// Driver Program to test above function
  
    let k = 2;
    let a = [ 1, 2, 3, 4 ];
    let n = a.length;
 
    let x = PointInKSquares(n, a, k);
    document.write("(" + x + ", " + x + ")");
 
// This code is contributed by Mayank Tyagi
 

输出：

(3, 3)

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