至少包含给定坐标一半的正方形的最小长度

给定二维平面中的N个点。任务是找到M的最小值，以使以2 * M边为原点的原点为中心的正方形在其内部或上方至少包含floor(N / 2)个点。

例子：

Input : N = 4
Points are: {(1, 2), (-3, 4), (1, 78), (-3, -7)}
Output : 4
The square with end point (4, 4), (-4, -4), (4, -4), (-4, 4) will contain the points (1, 2) and (-3, 4).
Smallest Possible value of M such that the square has at least 2 points is 4.

Input : N = 3
Points are: {(1, 2), (-3, 4), (1, 78)}
Output : 2
Square contains the point (1, 2). {floor(3/2) = 1}

方法：

1. 任何一点(x，y)的一个主要观察结果是，该点所在的最小M为max(abs(x)，abs(y))。
2. 使用点1，我们可以找到所有点的M的最小值并将其存储在数组中。
3. 对数组进行排序。
4. 现在，array [i]表示最小值M，这样，如果在边2 * M的正方形中需要i个点。 (因为i之下的所有点的M的最小值都小于或等于i)。
5. 打印array [floor(n / 2)]的值。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
  
#include 
using namespace std;
  
// Function to Calculate Absolute Value
int mod(int x)
{
    if (x >= 0)
        return x;
    return -x;
}
  
// Function to Calculate the Minimum value of M
void findSquare(int n)
{
    int points[n][2] = { { 1, 2 }, { -3, 4 }, 
                       { 1, 78 }, { -3, -7 } };
    int a[n];
  
    // To store the minimum M for each
    // point in array
    for (int i = 0; i < n; i++) {
        int x, y;
        x = points[i][0];
        y = points[i][1];
        a[i] = max(mod(x), mod(y));
    }
  
    // Sort the array
    sort(a, a + n);
  
    // Index at which atleast required point are
    // inside square of length 2*M
    int index = floor(n / 2) - 1;
    cout << "Minimum M required is: " << a[index] << endl;
}
  
// Driver Code
int main()
{
    int N;
    N = 4;
    findSquare(N);
  
    return 0;
}

Java

import java.util.*;
  
// Java program to find next identical year
class GFG
{
  
// Function to Calculate Absolute Value
static int mod(int x)
{
    if (x >= 0)
        return x;
    return -x;
}
  
// Function to Calculate the Minimum value of M
static void findSquare(int n)
{
    int points[][] = { { 1, 2 }, { -3, 4 }, 
                    { 1, 78 }, { -3, -7 } };
    int []a = new int[n];
  
    // To store the minimum M for each
    // point in array
    for (int i = 0; i < n; i++)
    {
        int x, y;
        x = points[i][0];
        y = points[i][1];
        a[i] = Math.max(mod(x), mod(y));
    }
  
    // Sort the array
    Arrays.sort(a);
  
    // Index at which atleast required point are
    // inside square of length 2*M
    int index = (int) (Math.floor(n / 2) - 1);
    System.out.println("Minimum M required is: " + a[index]);
}
  
// Driver Code
public static void main(String[] args) 
{
    int N;
    N = 4;
    findSquare(N);
}
}
  
// This code contributed by Rajput-Ji

Python3

# Python3 implementation of the
# above approach 
  
# Function to Calculate the 
# Minimum value of M 
def findSquare(n): 
  
    points = [[1, 2], [-3, 4], 
              [1, 78], [-3, -7]] 
    a = [None] * n 
  
    # To store the minimum M
    # for each point in array 
    for i in range(0, n): 
        x = points[i][0] 
        y = points[i][1] 
        a[i] = max(abs(x), abs(y)) 
      
    # Sort the array 
    a.sort() 
  
    # Index at which atleast required 
    # point are inside square of length 2*M 
    index = n // 2 - 1
    print("Minimum M required is:", a[index]) 
  
# Driver Code 
if __name__ == "__main__":
  
    N = 4
    findSquare(N)
      
# This code is contributed 
# by Rituraj Jain

C#

// C# program to find next identical year 
using System;
  
class GFG 
{ 
  
// Function to Calculate Absolute Value 
static int mod(int x) 
{ 
    if (x >= 0) 
        return x; 
    return -x; 
} 
  
// Function to Calculate the Minimum value of M 
static void findSquare(int n) 
{ 
    int [,]points = new int[4,2]{ { 1, 2 }, { -3, 4 }, 
                    { 1, 78 }, { -3, -7 } }; 
    int []a = new int[n]; 
  
    // To store the minimum M for each 
    // point in array 
    for (int i = 0; i < n; i++) 
    { 
        int x, y; 
        x = points[i,0]; 
        y = points[i,1]; 
        a[i] = Math.Max(mod(x), mod(y)); 
    } 
  
    // Sort the array 
    Array.Sort(a); 
  
    // Index at which atleast required point are 
    // inside square of length 2*M 
    int index = (int) (n / 2 - 1); 
    Console.WriteLine("Minimum M required is: " + a[index]); 
} 
  
// Driver Code 
public static void Main(String []args) 
{ 
    int N; 
    N = 4; 
    findSquare(N); 
} 
} 
  
// This code contributed by Arnab Kundu

PHP

= 0)
        return $x;
    return -$x;
}
  
// Function to Calculate the
// Minimum value of M
function findSquare($n)
{
    $points = array(array( 1, 2 ), 
                    array( -3, 4 ), 
                    array( 1, 78 ),
                    array( -3, -7 ));
    $a[$n] = array();
  
    // To store the minimum M for each
    // point in array
    for ($i = 0; $i < $n; $i++) 
    {
        $x; $y;
        $x = $points[$i][0];
        $y = $points[$i][1];
        $a[$i] = max(mod($x), mod($y));
    }
  
    // Sort the array
    sort($a);
  
    // Index at which atleast required point 
    // are inside square of length 2*M
    $index = floor($n / 2) - 1;
    echo "Minimum M required is: ", 
                  $a[$index], "\n";
}
  
// Driver Code
$N = 4;
findSquare($N);
  
// This code is contributed by ajit.
?>

输出：

Minimum M required is: 4