要删除剩余点在轴的一侧的最小点数

我们在笛卡尔平面中得到n个点。我们的任务是找到要删除的最小点数，以便使剩余点位于任何轴的一侧。

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 ：
这个问题是一个基于几何的构造性蛮力算法的简单示例。可以通过找到X轴和Y轴所有侧面上的点数来简单地找到该解决方案。最低限度就是答案。

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# rogram 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

输出 ：

1