要删除的最小点数以获得轴一侧的剩余点

我们在笛卡尔平面上有n个点。我们的任务是找到应该删除的最小点数，以便获得任何轴一侧的剩余点。

例子：

Input : 4
        1 1
        2 2
       -1 -1
       -2 2
Output : 1
Explanation :
If we remove (-1, -1) then all the remaining 
points are above x-axis. Thus the answer is 1.

Input : 3
        1 10
        2 3
        4 11
Output : 0
Explanation :
All points are already above X-axis. Hence the
answer is 0.

方法：
这个问题是几何上的建设性蛮力算法的一个简单例子。可以简单地通过查找 X 轴和 Y 轴所有边上的点数来接近该解决方案。这将是答案的最小值。

C++

// CPP program to find minimum points to be moved
// so that all points are on same side.
#include 
using namespace std;
typedef long long ll;
 
// Structure to store the coordinates of a point.
struct Point
{
    int x, y;
};
 
// Function to find the minimum number of points
int findmin(Point p[], int n)
{
    int a = 0, b = 0, c = 0, d = 0;
    for (int i = 0; i < n; i++)
    {
        // Number of points on the left of Y-axis.
        if (p[i].x <= 0)        
            a++;
 
        // Number of points on the right of Y-axis.
        else if (p[i].x >= 0)
            b++;
 
        // Number of points above X-axis.
        if (p[i].y >= 0)
            c++;
 
        // Number of points below X-axis.
        else if (p[i].y <= 0)
            d++;
    }
 
    return min({a, b, c, d});
}
 
// Driver Function
int main()
{
    Point p[] = { {1, 1}, {2, 2}, {-1, -1}, {-2, 2} };
    int n = sizeof(p)/sizeof(p[0]);
    cout << findmin(p, n);
    return 0;
}

Java

// Java program to find minimum points to be moved
// so that all points are on same side.
import java.util.*;
 
class GFG
{
 
// Structure to store the coordinates of a point.
static class Point
{
    int x, y;
 
    public Point(int x, int y)
    {
        this.x = x;
        this.y = y;
    }
};
 
// Function to find the minimum number of points
static int findmin(Point p[], int n)
{
    int a = 0, b = 0, c = 0, d = 0;
    for (int i = 0; i < n; i++)
    {
        // Number of points on the left of Y-axis.
        if (p[i].x <= 0)    
            a++;
 
        // Number of points on the right of Y-axis.
        else if (p[i].x >= 0)
            b++;
 
        // Number of points above X-axis.
        if (p[i].y >= 0)
            c++;
 
        // Number of points below X-axis.
        else if (p[i].y <= 0)
            d++;
    }
    return Math.min(Math.min(a, b),
                    Math.min(c, d));
}
 
// Driver Code
public static void main(String[] args)
{
    Point p[] = {new Point(1, 1), new Point(2, 2),
                 new Point(-1, -1), new Point(-2, 2)};
    int n = p.length;
    System.out.println(findmin(p, n));
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 program to find minimum points to be
# moved so that all points are on same side.
 
# Function to find the minimum number
# of points
def findmin(p, n):
 
    a, b, c, d = 0, 0, 0, 0
    for i in range(n):
         
        # Number of points on the left
        # of Y-axis.
        if (p[i][0] <= 0):    
            a += 1
 
        # Number of points on the right
        # of Y-axis.
        elif (p[i][0] >= 0):
            b += 1
 
        # Number of points above X-axis.
        if (p[i][1] >= 0):
            c += 1
 
        # Number of points below X-axis.
        elif (p[i][1] <= 0):
            d += 1
 
    return min([a, b, c, d])
 
# Driver Code
p = [ [1, 1], [2, 2], [-1, -1], [-2, 2] ]
n = len(p)
print(findmin(p, n))
     
# This code is contributed by Mohit Kumar

C#

// C# program to find minimum points to be moved
// so that all points are on same side.
using System;
     
class GFG
{
 
// Structure to store the coordinates of a point.
public class Point
{
    public int x, y;
 
    public Point(int x, int y)
    {
        this.x = x;
        this.y = y;
    }
};
 
// Function to find the minimum number of points
static int findmin(Point []p, int n)
{
    int a = 0, b = 0, c = 0, d = 0;
    for (int i = 0; i < n; i++)
    {
        // Number of points on the left of Y-axis.
        if (p[i].x <= 0)    
            a++;
 
        // Number of points on the right of Y-axis.
        else if (p[i].x >= 0)
            b++;
 
        // Number of points above X-axis.
        if (p[i].y >= 0)
            c++;
 
        // Number of points below X-axis.
        else if (p[i].y <= 0)
            d++;
    }
    return Math.Min(Math.Min(a, b),
                    Math.Min(c, d));
}
 
// Driver Code
public static void Main(String[] args)
{
    Point []p = {new Point(1, 1),
                 new Point(2, 2),
                 new Point(-1, -1),
                 new Point(-2, 2)};
    int n = p.Length;
    Console.WriteLine(findmin(p, n));
}
}
     
// This code is contributed by Princi Singh

Javascript

输出：