Minimum number of points to be removed to get remaining points on one side of axis

We are given n points in a Cartesian plane. Our task is to find the minimum number of points that should be removed in order to get the remaining points on one side of any axis.

Examples :

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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach :
This problem is a simple example of constructive brute force algorithm on Geometry. The solution can be approached simply by finding the number of points on all sides of the X-axis and Y-axis. The minimum of this will be the answer.

C++

// CPP program to find minimum points to be moved 
// so that all points are on same side. 
#include <bits/stdc++.h> 
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 

Output :

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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